\documentclass[12pt]{report} \usepackage{amsthm, amssymb, amsmath} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{0in} \setlength{\textheight}{8.5in} \setlength{\topmargin}{0in} \setlength{\headheight}{0in} \setlength{\headsep}{0in} \setlength{\parskip}{0pt} \setlength{\parindent}{20pt} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{problem}[theorem]{Example} \newtheorem{cor}[theorem]{Corollary} \def\CC{\mathbb{C}} \def\FF{\mathbb{F}} \def\GG{\mathbb{G}} \def\NN{\mathbb{N}} \def\QQ{\mathbb{Q}} \def\RR{\mathbb{R}} \def\ZZ{\mathbb{Z}} \def\calC{\mathcal{C}} \def\calE{\mathcal{E}} \def\calF{\mathcal{F}} \def\calG{\mathcal{G}} \def\calH{\mathcal{H}} \def\calL{\mathcal{L}} \def\catS{\mathcal{S}} \def\catU{\mathcal{U}} \def\alf{\alpha} \def\Lp{\left(} \def\Rp{\right)} \def\vt{\vec{t}} \def\vx{\vec{x}} \def\vy{\vec{y}} \def\be{\begin{enumerate}} \def\ee{\end{enumerate}} \def\beq{\begin{equation}} \def\eeq{\end{equation}} \def\map#1#2#3{#1\!:\!#2 \to #3} \def\ii{\item} \def\csum{\sum_{\mathrm{cyclic}}} \def\ssym{\sum_{\mathrm{sym}}} \newcounter{exc} \numberwithin{exc}{section} \newenvironment{exer}{\vspace{0.1in} \noindent \textbf{Problems for Section~\thesection} \vspace{0.1in} \begin{list}{\arabic{exc}.}{\usecounter{exc}}}{\end{list}} \begin{document} %% Title page \begin{titlepage} \begin{center} $A < B$ ($A$ is less than $B$) \vspace{1in} Kiran Kedlaya based on notes for the Math Olympiad Program (MOP) Version 1.0, last revised August 2, 1999 \vspace{1in} \copyright Kiran S. Kedlaya. This is an unfinished manuscript distributed for personal use only. In particular, any publication of all or part of this manuscript without prior consent of the author is strictly prohibited. \bigskip Please send all comments and corrections to the author at \texttt{kedlaya@math.mit.edu}. Thank you! \end{center} \end{titlepage} \chapter*{Introduction} These notes constitute a survey of the theory and practice of inequalities. While their intended audience is high-school students, primarily present and aspiring participants of the Math Olympiad Program (MOP), I hope they prove informative to a wider audience. In particular, those who experience inequalities via the Putnam competition, or via problem columns in such journals as \textit{Crux Mathematicorum} or the \textit{American Mathematical Monthly}, should find some benefit. Having named high-school students as my target audience, I must now turn around and admit that I have not made any effort to keep calculus out of the exposition, for several reasons. First, in certain places, rewriting to avoid calculus would make the exposition a lot more awkward. Second, the calculus I invoke is for the most part pretty basic, mostly properties of the first and second derivative. Finally, it is my experience that many Olympiad participants have studied calculus anyway. In any case, I have clearly flagged uses of calculus in the text, and I've included a crash course in calculus (Chapter~\ref{chap:calc}) to fill in the details. By no means is this primer a substitute for an honest treatise on inequalities, such as the \emph{magnum opus} of Hardy, Littlewood and P\'olya \cite{bib:ha} or its latter-day sequel \cite{bib:bmv}, nor for a comprehensive catalog of problems in the area, for which we have Stanley Rabinowitz' series \cite{bib:rab}. My aim, rather than to provide complete information, is to whet the reader's appetite for this beautiful and boundless subject. Also note that I have given geometric inequalities short shrift, except to the extent that they can be written in an algebraic or trigonometric form. ADD REFERENCE. Thanks to Paul Zeitz for his MOP 1995 notes, upon which these notes are ultimately based. (In particular, they are my source for symmetric sum notation.) Thanks also to the participants of the 1998 and 1999 MOPs for working through preliminary versions of these notes. \section*{Caveat solver!} It seems easier to fool oneself by constructing a false proof of an inequality than of any other type of mathematical assertion. All it takes is one reversed inequality to turn an apparently correct proof into a wreck. The adage ``if it seems too good to be true, it probably is'' applies in full force. To impress the gravity of this point upon the reader, we provide a little exercise in mathematical proofreading. Of the following X proofs, only Y are correct. Can you spot the fakes? PUT IN THE EXAMPLES. \chapter{Separable inequalities} This chapter covers what I call ``separable'' inequalities, those which can be put in the form \[ f(x_{1}) + \cdots + f(x_{n}) \geq c \] for suitably constrained $x_{1}, \dots, x_{n}$. For example, if one fixes the product, or the sum, of the variables, the AM-GM inequality takes this form, and in fact this will be our first example. \section{Smoothing, convexity and Jensen's inequality} The ``smoothing principle'' states that if you have a quantity of the form $f(x_{1}) + \cdots + f(x_{n})$ which becomes smaller as you move two of the variables closer together (while preserving some constraint, e.g.\ the sum of the variables), then the quantity is minimized by making the variables all equal. This remark is best illustrated with an example: the famous arithmetic mean and geometric mean (AM-GM) inequality. \begin{theorem}[AM-GM] Let $x_{1}, \dots, x_{n}$ be positive real numbers. Then \[ \frac{x_{1} + \cdots + x_{n}}{n} \geq \sqrt[n]{x_{1}\cdots x_{n}}, \] with equality if and only if $x_{1} = \cdots = x_{n}$. \end{theorem} \begin{proof} We will make a series of substitutions that preserve the left-hand side while strictly increasing the right-hand side. At the end, the $x_{i}$ will all be equal and the left-hand side will equal the right-hand side; the desired inequality will follow at once. (Make sure that you understand this reasoning before proceeding!) If the $x_{i}$ are not already all equal to their arithmetic mean, which we call $a$ for convenience, then there must exist two indices, say $i$ and $j$, such that $x_{i} < a < x_{j}$. (If the $x_{i}$ were all bigger than $a$, then so would be their arithmetic mean, which is impossible; similarly if they were all smaller than $a$.) We will replace the pair $x_{i}$ and $x_{j}$ by \[ x_{i}' = a, x_{j}' = x_{i} + x_{j} - a; \] by design, $x_{i}'$ and $x_{j}'$ have the same sum as $x_{i}$ and $x_{j}$, but since they are closer together, their product is larger. To be precise, \[ a(x_{i} + x_{j} - a) = x_{i}x_{j} + (x_{j} - a)(a - x_{i}) > x_{i}x_{j} \] because $x_{j}-a$ and $a-x_{i}$ are positive numbers. By this replacement, we increase the number of the $x_{i}$ which are equal to $a$, preserving the left-hand side of the desired inequality by increasing the right-hand side. As noted initially, eventually this process ends when all of the $x_{i}$ are equal to $a$, and the inequality becomes equality in that case. It follows that in all other cases, the inequality holds strictly. \end{proof} Note that we made sure that the replacement procedure terminates in a finite number of steps. If we had proceeded more naively, replacing a pair of $x_i$ by \emph{their} arithmetic meaon, we would get an infinite procedure, and then would have to show that the $x_i$ were ``converging'' in a suitable sense. (They do converge, but making this precise would require some additional effort which our alternate procedure avoids.) A strong generalization of this smoothing can be formulated for an arbitrary convex function. Recall that a set of points in the plane is said to be convex if the line segment joining any two points in the set lies entirely within the set. A function $f$ defined on an interval (which may be open, closed or infinite on either end) is said to be \emph{convex} if the set \[ \{ (x, y) \in \RR^{2}: y \geq f(x) \} \] is convex. We say $f$ is \emph{concave} if $-f$ is convex. (This terminology was standard at one time, but today most calculus textbooks use ``concave up'' and ``concave down'' for our ``convex'' and ``concave''. Others use the evocative sobriquets ``holds water'' and ``spills water''.) A more enlightening way to state the definition might be that $f$ is convex if for any $t \in [0,1]$ and any $x,y$ in the domain of $f$, \[ tf(x) + (1-t)f(y) \geq f(tx + (1-t)y). \] If $f$ is continuous, it suffices to check this for $t=1/2$. Conversely, a convex function is automatically continuous except possibly at the endpoints of the interval on which it is defined. DIAGRAM. \begin{theorem} If $f$ is a convex function, then the following statements hold: \begin{enumerate} \ii If $a \leq b < c \leq d$, then $\frac{f(c)-f(a)}{c-a} \leq \frac{f(d)-f(b)}{d-b}$. (The slopes of secant lines through the graph of $f$ increase with either endpoint.) \ii If $f$ is differentiable everywhere, then its derivative (that is, the slope of the tangent line to the graph of $f$ is an increasing function.) \end{enumerate} \end{theorem} The utility of convexity for proving inequalities comes from two factors. The first factor is Jensen's inequality, which one may regard as a formal statement of the ``smoothing principle'' for convex functions. \begin{theorem}[Jensen] Let $f$ be a convex function on an interval $I$ and let $w_{1}, \dots, w_{n}$ be nonnegative real numbers whose sum is 1. Then for all $x_{1}, \dots, x_{n} \in I$, \[ w_{1}f(x_{1}) + \cdots + w_{n} f(x_{n}) \geq f(w_{1}x_{1} + \cdots + w_{n}x_{n}). \] \end{theorem} \begin{proof} An easy induction on $n$, the case $n=2$ being the second definition above. \end{proof} The second factor is the ease with which convexity can be checked using calculus, namely via the second derivative test. \begin{theorem} Let $f$ be a twice-differentiable function on an open interval $I$. Then $f$ is convex on $I$ if and only if $f''(x) \geq 0$ for all $x \in I$. \end{theorem} For example, the AM-GM inequality can be proved by noting that $f(x) = \log x$ is concave; its first derivative is $1/x$ and its second $-1/x^{2}$. In fact, one immediately deduces a weighted AM-GM inequality; as we will generalize AM-GM again later, we will not belabor this point. We close this section by pointing out that separable inequalities sometimes concern functions which are not necessarily convex. Nonetheless one can prove something! \begin{problem}[USA, 1998] Let $a_0, \dots, a_n$ be numbers in the interval $(0, \pi/2)$ such that \[ \tan (a_0 - \pi/4) + \tan (a_1 - \pi/4) + \cdots + \tan(a_n - \pi/4) \geq n-1. \] Prove that $\tan a_0 \tan a_1 \cdots \tan a_n \geq n^{n+1}$. \end{problem} \begin{proof}[Solution.] Let $x_i = \tan (a_i - \pi/4)$ and $y_i = \tan a_i = (1+x_i)/(1-x_i)$, so that $x_i \in (-1,1)$. The claim would follow immediately from Jensen's inequality if only the function $f(x) = \log(1+x)/(1-x)$ were convex on the interval $(-1,1)$, but alas, it isn't. It's concave on $(-1,0]$ and convex on $[0,1)$. So we have to fall back on the smoothing principle. What happens if we try to replace $x_i$ and $x_j$ by two numbers that have the same sum but are closer together? The contribution of $x_i$ and $x_j$ to the left side of the desired inequality is \[ \frac{1+x_i}{1-x_i} \cdot \frac{1+x_j}{1-x_j} = 1 + \frac{2}{\frac{x_ix_j+1}{x_i+x_j} - 1}. \] The replacement in question will increase $x_ix_j$, and so will decrease the above quantity provided that $x_i + x_j > 0$. So all we need to show is that we can carry out the smoothing process so that every pair we smooth satisfies this restriction. Obviously there is no problem if all of the $x_i$ are positive, so we concentrate on the possibility of having $x_i < 0$. Fortunately, we can't have more than one negative $x_i$, since $x_0 + \cdots + x_n \geq n-1$ and each $x_i$ is less than 1. (So if two were negative, the sum would be at most the sum of the remaining $n-1$ terms, which is less than $n-1$.) Moreover, if $x_0 < 0$, we could not have $x_0 + x_j < 0$ for $j=1, \dots, n$, else we would have the contradiction \[ x_0 + \cdots + x_n \leq (1-n)x_0 < n-1. \] Thus $x_0 + x_j > 0$ for some $j$, and we can replace these two by their arithmetic mean. Now all of the $x_i$ are positive and smoothing (or Jensen) may continue without further restrictions, yielding the desired inequality. \end{proof} \begin{exer} \ii Make up a problem by taking a standard property of convex functions, and specializing to a particular function. The less evident it is where your problem came from, the better! \ii Given real numbers $x_{1}, \dots, x_{n}$, what is the minimum value of \[ |x-x_{1}| + \cdots + |x-x_{n}|? \] \ii (via Titu Andreescu) If $f$ is a convex function and $x_1, x_2, x_3$ lie in its domain, then \begin{eqnarray*} \lefteqn{f(x_1) + f(x_2) + f(x_3) + f\left( \frac{x_1+x_2+x_3}{3}\right)} && \\ &\geq& \frac 43 \left[ f\left(\frac{x_1+x_2}{2} \right) + f\left(\frac{x_2+x_3}{2} \right) \right. \\ && \left. + f\left(\frac{x_3+x_1}{2} \right) \right]. \end{eqnarray*} \ii (USAMO 1974/1) For $a,b,c>0$, prove that $a^ab^bc^c \geq (abc)^{(a+b+c)/3}$. \ii (India, 1995) Let $x_{1}, \dots, x_{n}$ be $n$ positive numbers whose sum is 1. Prove that \[ \frac{x_{1}}{\sqrt{1-x_{1}}} + \cdots + \frac{x_{n}}{\sqrt{1-x_{n}}} \geq \sqrt{\frac{n}{n-1}}. \] \ii Let $A,B,C$ be the angles of a triangle. Prove that \be %\ii BLAH (i.e. circumradius is at least twice inradius) \ii $\sin A + \sin B + \sin C \leq 3\sqrt{3}/2$; \ii $\cos A + \cos B + \cos C \leq 3/2$; \ii $\sin A/2 \sin B/2 \sin C/2 \leq 1/8$; \ii $\cot A + \cot B + \cot C \geq \sqrt{3}$ (i.e. the Brocard angle is at most $\pi/6$). \ee (Beware: not all of the requisite functions are convex everywhere!) \ii (Vietnam, 1998) Let $x_{1}, \dots, x_{n}$ ($n \geq 2$) be positive numbers satisfying \[ \frac{1}{x_{1}+1998} + \frac{1}{x_{2}+1998} + \cdots + \frac{1}{x_{n}+1998} = \frac{1}{1998}. \] Prove that \[ \frac{\sqrt[n]{x_{1}x_{2}\cdots x_{n}}}{n-1} \geq 1998. \] (Again, beware of nonconvexity.) \ii Let $x_1, x_2, \dots$ be a sequence of positive real numbers. If $a_k$ and $g_k$ are the arithmetic and geometric means, respectively, of $x_1, \dots, x_k$, prove that \begin{align} \frac{a_n^n}{g_k^n} &\geq \frac{a_{n-1}^{n-1}}{g_k^{n-1}} \\ n(a_n - g_n) &\geq (n-1)(a_{n-1}-g_{n-1}). \end{align} These strong versions of the AM-GM inequality are due to Rado \cite[Theorem~60]{bib:ha} and Popoviciu \cite{bib:pop}, respectively. (These are just a sample of the many ways the AM-GM inequality can be sharpened, as evidenced by \cite{bib:bmv}.) \end{exer} \section{Unsmoothing and noninterior extrema} A convex function has no interior local maximum. (If it had an interior local maximum at $x$, then the secant line through $(x-\epsilon, f(x-\epsilon))$ and $(x + \epsilon, f(x+\epsilon))$ would lie under the curve at $x$, which cannot occur for a convex function.) Even better, a function which is \emph{linear} in a given variable, as the others are held fixed, attains no extrema in either direction except at its boundary values. \begin{exer} \ii (IMO 1974/5) Determine all possible values of \[ S= \frac{a}{a+b+d} + \frac{b}{a + b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}\] where $a,b,c,d$ are arbitrary positive numbers. \ii (Bulgaria, 1995) Let $n \geq 2$ and $0 \leq x_{i} \leq 1$ for all $i=1,2,\dots, n$. Show that \[ (x_{1} + x_{2} + \cdots + x_{n}) - (x_{1}x_{2} + x_{2}x_{3} + \cdots + x_{n}x_{1}) \leq \left\lfloor \frac{n}{2} \right\rfloor, \] and determine when there is equality. \end{exer} %\section{Convexity and trigonometric functions} \section{Discrete smoothing} The notions of smoothing and convexity can also be applied to functions only defined on integers. \begin{problem} How should $n$ balls be put into $k$ boxes to minimize the number of pairs of balls which lie in the same box? \end{problem} \begin{proof}[Solution] In other words, we want to minimize $\sum_{i=1}^{k} \binom{n_{i}}{2}$ over sequences $n_{1}, \dots, n_{k}$ of nonnegative integers adding up to $n$. If $n_{i} - n_{j} \geq 2$ for some $i,j$, then moving one ball from $i$ to $j$ decreases the number of pairs in the same box by \[ \binom{n_{i}}{2} - \binom{n_{i}-1}{2} + \binom{n_{j}}{2} - \binom{n_{j}+1}{2} = n_{i} - n_{j} - 1 > 0. \] Thus we minimize the number of pairs by making sure no two boxes differ by more than one ball; one can easily check that the boxes must each contain $\lfloor n/k\rfloor$ or $\lceil n/k \rceil$ balls. \end{proof} \begin{exer} \ii (Germany, 1995) Prove that for all integers $k$ and $n$ with $1 \leq k \leq 2n$, \[ \binom{2n+1}{k-1} + \binom{2n+1}{k+1} \geq 2 \cdot \frac{n+1}{n+2} \cdot \binom{2n+1}{k}. \] \ii (Arbelos) Let $a_{1}, a_{2}, \dots$ be a convex sequence of real numbers, which is to say $a_{k-1}+a_{k+1} \geq 2a_{k}$ for all $k \geq 2$. Prove that for all $n \geq 1$, \[ \frac{a_{1} + a_{3}+ \cdots + a_{2n+1}}{n+1} \geq \frac{a_{2} + a_{4}+ \cdots + a_{2n}}{n}. \] \ii (USAMO 1993/5) Let $a_{0}, a_{1}, a_{2}, \dots$ be a sequence of positive real numbers satisfying $a_{i-1}a_{i+1} \leq a_{i}^{2}$ for $i=1,2,3,\dots$. (Such a sequence is said to be \emph{log concave}.) Show that for each $n \geq 1$, \[ \frac{a_{0} + \cdots + a_{n}}{n+1} \cdot \frac{a_{1} + \cdots + a_{n-1}}{n-1} \geq \frac{a_{0} + \cdots + a_{n-1}}{n} \cdot \frac{a_{1} + \cdots + a_{n}}{n}. \] \ii (MOP 1997) Given a sequence $\{x_n\}_{n=0}^\infty$ with $x_n > 0$ for all $n\geq0$, such that the sequence $\{a^n x_n\}_{n=0}^\infty$ is convex for all $a>0$, show that the sequence $\{\log x_n\}_{n=0}^\infty$ is also convex. \ii How should the numbers $1, \dots, n$ be arranged in a circle to make the sum of the products of pairs of adjacent numbers as large as possible? As small as possible? \end{exer} \chapter{Symmetric polynomial inequalities} This section is a basic guide to polynomial inequalities, which is to say, those inequalities which are in (or can be put into) the form $P(x_{1}, \dots, x_{n}) \geq 0$ with $P$ a symmetric polynomial in the real (or sometimes nonnegative) variables $x_{1}, \dots, x_{n}$. \section{Introduction to symmetric polynomials} Many inequalities express some symmetric relationship between a collection of numbers. For this reason, it seems worthwhile to brush up on some classical facts about symmetric polynomials. For arbitrary $x_{1}, \dots, x_{n}$, the coefficients $c_{1}, \dots, c_{n}$ of the polynomial \[ (t + x_{1})\cdots (t + x_{n}) = t^{n} + c_{1} t^{n-1} + \cdots + c_{n-1}t + c_{n} \] are called the \emph{elementary symmetric functions} of the $x_{i}$. (That is, $c_{k}$ is the sum of the products of the $x_{i}$ taken $k$ at a time.) Sometimes it proves more convenient to work with the \emph{symmetric averages} \[ d_{i} = \frac{c_{i}}{\binom ni}. \] For notational convenience, we put $c_{0} = d_{0} = 1$ and $c_{k} = d_{k} = 0$ for $k > n$. Two basic inequalities regarding symmetric functions are the following. (Note that the second follows from the first.) \begin{theorem}[Newton] If $x_1, \dots, x_n$ are nonnegative, then %% real?? \[ d_i^2 \geq d_{i-1} d_{i+1} \qquad i = 1, \dots, n-1. \] \end{theorem} \begin{theorem}[Maclaurin] If $x_{1}, \dots, x_{n}$ are positive, then \[ d_{1} \geq d_{2}^{1/2} \geq \cdots \geq d_{n}^{1/n}, \] with equality if and only if $x_{1} = \cdots = x_{n}$. \end{theorem} These inequalities and many others can be proved using the following trick. \begin{theorem} \label{trick} Suppose the inequality \[ f(d_{1}, \dots, d_{k}) \geq 0 \] holds for all real (resp. positive) $x_{1}, \dots, x_{n}$ for some $n \geq k$. Then it also holds for all real (resp. positive) $x_{1}, \dots, x_{n+1}$. \end{theorem} \begin{proof} Let \[ P(t) = (t + x_{1})\cdots (t+x_{n+1}) = \sum_{i=0}^{n+1} \binom{n+1}{i} d_{i} t^{n+1-i} \] be the monic polynomial with roots $-x_{1}, \dots, -x_{n}$. Recall that between any two zeros of a differentiable function, there lies a zero of its derivative (Rolle's theorem). Thus the roots of $P'(t)$ are all real if the $x_{i}$ are real, and all negative if the $x_{i}$ are positive. Since \[ P'(t) = \sum_{i=0}^{n+1} (n+1-i) \binom{n+1}{i} d_{i} t^{n-i} = (n+1) \sum_{i=0}^{n} \binom{n}{i} d_{i} t^{n-i}, \] we have by assumption $f(d_{1}, \dots, d_{k}) \geq 0$. \end{proof} Incidentally, the same trick can also be used to prove certain polynomial identities. \begin{exer} \ii Prove that every symmetric polynomial in $x_{1}, \dots, x_{n}$ can be (uniquely) expressed as a polynomial in the elementary symmetric polynomials. \ii Prove Newton's and Maclaurin's inequalities. \ii Prove Newton's identities: if $s_{i} = x_{1}^{i} + \cdots + x_{n}^{i}$, then \[ c_{0} s_{k} + c_{1} s_{k-1} + \cdots + c_{k-1} s_{1} + k c_{k} = 0. \] (Hint: first consider the case $n=k$.) \ii (Hungary) Let $f(x)=x^n+a_{n-1}x^{n-1}+\cdots +a_1x+1$ be a polynomial with non-negative real coefficients and $n$ real roots. Prove that $f(x) \geq (1+x)^n$ for all $x\geq 0$. \ii (Gauss-??) Let $P(z)$ be a polynomial with complex coefficients. Prove that the (complex) zeroes of $P'(z)$ all lie in the convex hull of the zeroes of $P(z)$. Deduce that if $S$ is a convex subset of the complex plane (e.g., the unit disk), and if $\Re f(d_{1}, \dots, d_{k}) \geq 0$ holds for all $x_{1}, \dots, x_{n} \in S$ for some $n \geq k$, then the same holds for $x_{1}, \dots, x_{n+1} \in S$. \ii Prove Descartes' Rule of Signs: let $P(x)$ be a polynomial with real coefficients written as $P(x) = \sum a_{k_{i}} x^{k_{i}}$, where all of the $a_{k_{i}}$ are nonzero. Prove that the number of positive roots of $P(x)$, counting multiplicities, is equal to the number of sign changes (the number of $i$ such that $a_{k_{i-1}}a_{k_{i}} < 0$) minus a nonnegative even integer. (For negative roots, apply the same criterion to $P(-x)$.) \end{exer} \section{The idiot's guide to homogeneous inequalities} Suppose one is given a homogeneous symmetric polynomial $P$ and asked to prove that $P(x_{1}, \dots, x_{n}) \geq 0$. How should one proceed? Our first step is purely formal, but may be psychologically helpful. We introduce the following notation: \[ \ssym Q(x_{1}, \dots, x_{n}) = \sum_{\sigma} Q(x_{\sigma(1)}, \dots, x_{\sigma(n)}) \] where $\sigma$ runs over all permutations of $1, \dots, n$ (for a total of $n!$ terms). For example, if $n = 3$, and we write $x,y,z$ for $x_{1}, x_{2}, x_{3}$, \begin{align*} \ssym x^{3} &= 2x^{3} + 2y^{3} + 2z^{3} \\ \ssym x^{2}y &= x^{2}y + y^{2}z + z^{2}x + x^{2}z + y^{2}x + z^{2}y \\ \ssym xyz &= 6xyz. \end{align*} Using symmetric sum notation can help prevent errors, particularly when one begins with rational functions whose denominators must first be cleared. Of course, it is always a good algebra check to make sure that equality continues to hold when it's supposed to. In passing, we note that other types of symmetric sums can be useful when the inequalities in question do not have complete symmetry, most notably cyclic summation \[ \csum x^{2}y = x^{2}y + y^{2}z + z^{2}x. \] However, it is probably a bad idea to mix, say, cyclic and symmetric sums in the same calculation! Next, we attempt to ``bunch'' the terms of $P$ into expressions which are nonnegative for a simple reason. For example, \[ \ssym (x^{3} - xyz) \geq 0 \] by the AM-GM inequality, while \[ \ssym (x^{2}y^{2} - x^{2}yz) \geq 0 \] by a slightly jazzed-up but no more sophisticated argument: we have $x^{2}y^{2} + x^{2}z^{2} \geq x^{2}yz$ again by AM-GM. We can formalize what we are doing using the notion of majorization. If $s = (s_{1}, \dots, s_{n})$ and $t = (t_{1}, \dots, t_{n})$ are two nonincreasing sequences, we say that $s$ \emph{majorizes} $t$ if $s_{1} + \cdots + s_{n} = t_{1} + \cdots + t_{n}$ and $s_{1} + \cdots + s_{i} \geq t_{1} + \cdots + t_{i}$ for $i = 1, \dots, n$. \begin{theorem}[``Bunching''] If $s$ and $t$ are sequences of nonnegative reals such that $s$ majorizes $t$, then \[ \ssym x_{1}^{s_{1}} \cdots x_{n}^{s_{n}} \geq \ssym x_{1}^{t_{1}} \cdots x_{n}^{t_{n}}. \] \end{theorem} \begin{proof} One first shows that if $s$ majorizes $t$, then there exist nonnegative constants $k_{\sigma}$, as $\sigma$ ranges over the permutations of $\{1, \dots, n\}$, whose sum is 1 and such that \[ \sum_\sigma k_{\sigma} (s_{\sigma_1}, \dots, s_{\sigma_n}) = (t_1, \dots, t_n) \] (and conversely). Then apply weighted AM-GM as follows: \begin{eqnarray*} \sum_{\sigma} x_1^{s_{\sigma(n)}} \cdots x_n^{s_{\sigma(n)}} &=& \sum_{\sigma,\tau} k_{\tau} x_1^{s_{\sigma(\tau(1))}} \cdots x_n^{s_{\sigma(\tau(n))}} \\ &\geq& \sum_{\sigma} x_1^{\sum_\tau k_\tau s_{\sigma(\tau(1))}} \cdots x_n^{\sum_\tau k_\tau s_{\sigma(\tau(n))}}\\ &=& \sum_{\sigma} x_1^{t_{\sigma(1)}} \cdots x_n^{t_{\sigma(n)}}. \end{eqnarray*} \end{proof} If the indices in the above proof are too bewildering, here's an example to illustrate what's going on: for $s = (5,2,1)$ and $t = (3,3,2)$, we have \[ (3,3,2) = (5,2,1) + \] and so BLAH. %More generally, this argument shows that if %$f(t_{1}, \dots, t_{n})$ is a convex symmetric function and %$s$ majorizes $t$, then $f(s) \geq f(t)$. (See Section LATER %for the definition of convexity of a multivariate function.) \begin{problem}[USA, 1997] Prove that, for all positive real numbers $a,b,c$, \[ \frac{1}{a^{3}+b^{3}+abc} + \frac{1}{b^{3}+c^{3}+abc} + \frac{1}{c^{3}+a^{3}+abc} \leq \frac{1}{abc}. \] \end{problem} \begin{proof}[Solution] Clearing denominators and multiplying by 2, we have \[ \ssym (a^{3}+b^{3}+abc)(b^{3}+c^{3}+abc)abc \leq 2(a^{3}+b^{3}+abc)(b^{3}+c^{3}+abc)(c^{3}+a^{3}+abc), \] which simplifies to \[ \ssym a^{7}bc + 3 a^{4}b^{4}c + 4 a^{5}b^{2}c^{2} + a^{3}b^{3}c^{3} \leq \ssym a^{3}b^{3}c^{3} + 2 a^{6}b^{3} + 3 a^{4}b^{4}c + 2a^{5}b^{2}c^{2} + a^{7}bc, \] and in turn to \[ \ssym 2 a^{6}b^{3} - 2 a^{5}b^{2}c^{2} \geq 0, \] which holds by bunching. \end{proof} In this case we were fortunate that after slogging through the algebra, the resulting symmetric polynomial inequality was quite straightforward. Alas, there are cases where bunching will not suffice, but for these we have the beautiful inequality of Schur. Note the extra equality condition in Schur's inequality; this is a much stronger result than AM-GM, and so cannot follow from a direct AM-GM argument. In general, one can avoid false leads by remembering that all of the steps in the proof of a given inequality must have equality conditions at least as permissive as those of the desired result! \begin{theorem}[Schur] Let $x, y, z$ be nonnegative real numbers. Then for any $r>0$, \[ x^{r} (x-y)(x-z) + y^{r}(y-z)(y-x) + z^{r}(z-x)(z-y) \geq 0. \] Equality holds if and only if $x=y=z$, or if two of $x,y,z$ are equal and the third is zero. \end{theorem} \begin{proof} Since the inequality is symmetric in the three variables, we may assume without loss of generality that $x \geq y \geq z$. Then the given inequality may be rewritten as \[ (x-y)[x^{r}(x-z) - y^{r}(y-z)] + z^{r}(x-z)(y-z) \geq 0, \] and every term on the left-hand side is clearly nonnegative. \end{proof} Keep an eye out for the trick we just used: creatively rearranging a polynomial into the sum of products of obviously nonnegative terms. The most commonly used case of Schur's inequality is $r=1$, which, depending on your notation, can be written \[ 3d_{1}^{3} + d_{3} \geq 4d_{1}d_{2} \quad \mbox{or} \quad \ssym x^{3} - 2x^{2}y + xyz \geq 0. \] If Schur is invoked with no mention $r$, you should assume $r=1$. \begin{problem} (Japan, 1997) \item (Japan, 1997) Let $a,b,c$ be positive real numbers, Prove that \[ \frac{(b+c-a)^2}{(b+c)^2+a^2} + \frac{(c+a-b)^2}{(c+a)^2+b^2} + \frac{(a+b-c)^2}{(a+b)^2+c^2} \geq \frac 35. \] \end{problem} \begin{proof}[Solution] We first simplify slightly, and introduce symmetric sum notation: \[ \ssym \frac{2ab + 2ac}{a^{2}+b^{2}+c^{2}+2bc} \leq \frac{24}{5}. \] Writing $s = a^{2}+b^{2}+c^{2}$, and clearing denominators, this becomes \[ 5 s^{2} \ssym ab + 10 s \ssym a^{2}bc + 20 \ssym a^{3}b^{2}c \leq 6 s^{3} + 6 s^{2} \ssym ab + 12 s \ssym a^{2}bc + 48 a^{2}b^{2}c^{2} \] which simplifies a bit to \[ 6s^{3} + s^{2} \ssym ab + 2s \ssym a^2bc + 8 \ssym a^{2}b^{2}c^{2} \geq 10 s \ssym a^{2}bc + 20 \ssym a^{3}b^{2}c. \] Now we multiply out the powers of $s$: \[ \ssym 3a^{6} + 2a^{5}b -2 a^{4}b^{2} + 3a^{4}bc + 2a^{3}b^{3} - 12 a^{3}b^{2}c + 4 a^{2}b^{2}c^{2} \geq 0. \] The trouble with proving this by AM-GM alone is the $a^{2}b^{2}c^{2}$ with a \emph{positive} coefficient, since it is the term with the most evenly distributed exponents. We save face using Schur's inequality (multiplied by $4abc$:) \[ \ssym 4a^{4}bc - 8a^{3}b^{2}c + 4a^{2}b^{2}c^{2} \geq 0, \] which reduces our claim to \[ \ssym 3a^{6} + 2a^{5}b -2 a^{4}b^{2} - a^{4}bc + 2a^{3}b^{3} - 4 a^{3}b^{2}c \geq 0. \] Fortunately, this is a sum of four expressions which are nonnegative by weighted AM-GM: \begin{eqnarray*} 0 &\leq& 2\ssym (2a^6 + b^6)/3 - a^4b^2 \\ 0 &\leq& \ssym (4a^6+b^6+c^6)/6 - a^4bc \\ 0 &\leq& 2\ssym (2a^3b^3+c^3a^3)/3 - a^3b^2c \\ 0 &\leq& 2 \ssym (2a^5b+a^5c+ab^5+ac^5)/6 - a^3b^2c. \end{eqnarray*} Equality holds in each case of AM-GM, and in Schur, if and only if $a=b=c$. \end{proof} \begin{exer} \ii Suppose $r$ is an even integer. Show that Schur's inequality still holds when $x,y,z$ are allowed to be arbitrary real numbers (not necessarily positive). \ii (Iran, 1996) Prove the following inequality for positive real numbers $x,y,z$: \[ (xy + yz + zx) \left( \frac{1}{(x+y)^{2}} + \frac{1}{(y+z)^{2}} + \frac{1}{(z+x)^{2}} \right) \geq \frac 94. \] \ii The author's solution to the USAMO 1997 problem also uses bunching, but in a subtler way. Can you find it? (Hint: try replacing each summand on the left with one that factors.) \ii (MOP 1998) \ii Prove that for $x,y,z > 0$, \[ \frac{x}{(x+y)(x+z)} + \frac{y}{(y+z)(y+x)} + \frac{z}{(z+x)(z+y)} \leq \frac{9}{4(x+y+z)}. \] \end{exer} \section{Variations: inhomogeneity and constraints} One can complicate the picture from the previous section in two ways: one can make the polynomials inhomogeneous, or one can add additional constraints. In many cases, one can reduce to a homogeneous, unconstrained inequality by creative manipulations or substitutions; we illustrate this process here. \begin{problem}[IMO 1995/2] \label{imo1995} Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that \[ \frac{1}{a^{3}(b+c)} + \frac{1}{b^{3}(c+a)} + \frac{1}{c^{3}(a+b)} \geq \frac{3}{2}. \] \end{problem} \begin{proof}[Solution.] We eliminate both the nonhomogeneity and the constraint by instead proving that \[ \frac{1}{a^{3}(b+c)} + \frac{1}{b^{3}(c+a)} + \frac{1}{c^{3}(a+b)} \geq \frac{3}{2(abc)^{4/3}}. \] This still doesn't look too appetizing; we'd prefer to have simpler denominators. So we make the additional substitution $a = 1/x, b = 1/y, c = 1/z$ $a = x/y$, $b = y/z$, $c = z/x$, in which case the inequality becomes \begin{equation} \label{imo95} \frac{x^{2}}{y+z} + \frac{y^{2}}{z+x} + \frac{z^{2}}{x+y} \geq \frac{3}{2}(xyz)^{1/3}. \end{equation} Now we may follow the paradigm: multiply out and bunch. We leave the details to the reader. (We will revisit this inequality several times later.) \end{proof} On the other hand, sometimes more complicated maneuvers are required, as in this offbeat example. \begin{problem}[Vietnam, 1996] Let $a,b,c,d$ be four nonnegative real numbers satisfying the conditions \[ 2(ab+ac+ad+bc+bd+cd) + abc + abd + acd + bcd = 16. \] Prove that \[ a+b+c+d \geq \frac 23 (ab+ac+ad+bc+bd+cd) \] and determine when equality holds. \end{problem} \begin{proof}[Solution (by Sasha Schwartz).] Adopting the notation from the previous section, we want to show that \[ 3d_{2} + d_{3} = 4 \implies d_{1} \geq d_{2}. \] Assume on the contrary that we have $3d_2 + d_3 = 4$ but $d_1 < d_2$. By Schur's inequality plus Theorem~\ref{trick}, we have \[ 3d_1^3 + d_3 \geq 4d_1 d_2. \] Substituting $d_3 = 4-3d_2$ gives \[ 3d_1^3 + 4 \geq (4d_1 + 3)d_2 > 4d_1^2 + 3d_1, \] which when we collect and factor implies $(3d_1-4)(d_1^2-1) > 0$. However, on one hand $3d_1 - 4 < 3d_2 - 4 = -d_3 < 0$; on the other hand, by Maclaurin's inequality $d_1^2 \geq d_2 > d_1$, so $d_1 > 1$. Thus $(3d_1 - 4)(d_1^2-1)$ is negative, a contradiction. As for equality, we see it implies $(3d_1 -4)(d_1^2-1) = 0$ as well as equality in Maclaurin and Schur, so $d_1 = d_2 = d_3 = 1$. %Of course equality should occur when all of the variables are equal; %one checks that the only possible common value is $a=b=c=d=1$. \end{proof} \begin{exer} \ii (IMO 1984/1) Prove that $0\leq yz+zx+xy-2xyz\leq 7/27$, where $x,y$ and $z$ are non-negative real numbers for which $x+y+z=1$. \ii (Ireland, 1998) Let $a,b,c$ be positive real numbers such that $a+b+c \geq abc$. Prove that $a^2+b^2+c^2 \geq abc$. (In fact, the right hand side can be improved to $\sqrt{3}abc$.) \ii (Bulgaria, 1998) Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that \[ \frac{1}{1+a+b} + \frac{1}{1+b+c} + \frac{1}{1+c+a} \leq \frac{1}{2+a} + \frac{1}{2+b} + \frac{1}{2+c}. \] \end{exer} \section{Substitutions, algebraic and trigonometric} Sometimes a problem can be simplified by making a suitable substition. In particular, this technique can often be used to simplify unwieldy constraints. One particular substitution occurs frequently in problems of geometric origin. The condition that the numbers $a,b,c$ are the sides of a triangle is equivalent to the constraints \[ a+b > c, \quad b+c >a, \quad c+a >b \] coming from the triangle inequality. If we let $x = (b+c-a)/2, y = (c+a-b)/2, z= (a+b-c)/2$, then the constraints become $x,y,z > 0$, and the original variables are also easy to express: \[ a = y+z, \quad b = z+x, \quad c = x+y. \] %EXAMPLE. A more exotic substitution can be used when the constraint $a+b+c=abc$ is given. Put \[ \alpha = \arctan a, \quad \beta = \arctan b, \quad \gamma = \arctan c; \] then $\alpha + \beta + \gamma$ is a multiple of $\pi$. (If $a,b,c$ are positive, one can also write them as the cotangents of three angles summing to $\pi/2$.) \begin{exer} \ii (IMO 1983/6) Let $a,b,c$ be the lengths of the sides of a triangle. Prove that \[a^2b(a-b)+b^2 c(b-c) + c^2a(c-a) \geq 0,\] and determine when equality occurs. \ii (Asian Pacific, 1996) Let $a,b,c$ be the lengths of the sides of a triangle. Prove that \[ \sqrt{a+b-c} + \sqrt{b+c-a} + \sqrt{c+a-b} \leq \sqrt{a} + \sqrt{b} + \sqrt{c}. \] \ii (MOP, 1999) Let $a,b,c$ be lengths of the the sides of a triangle. Prove that \[ a^3 + b^3 + c^3 - 3abc \geq 2 \max \{|a-b|^3, |b-c|^3, |c-a|^3\}. \] \ii (Arbelos) Prove that if $a,b,c$ are the sides of a triangle and \[ 2(ab^{2}+bc^{2}+ca^{2}) = a^{2}b+b^{2}c+c^{2}a + 3abc, \] then the triangle is equilateral. \ii (Korea, 1998) For positive real numbers $a,b,c$ with $a+b+c=abc$, show that \[ \frac{1}{\sqrt{1+a^{2}}} + \frac{1}{\sqrt{1+b^{2}}} + \frac{1}{\sqrt{1+c^{2}}} \leq \frac{3}{2}, \] and determine when equality occurs. (Try proving this by dehomogenizing and you'll appreciate the value of the trig substitution!) \end{exer} \chapter{The toolbox} In principle, just about any inequality can be reduced to the basic principles we have outlined so far. This proves to be fairly inefficient in practice, since once spends a lot of time repeating the same reduction arguments. More convenient is to invoke as necessary some of the famous classical inequalities described in this chapter. We only barely scratch the surface here of the known body of inequalities; already \cite{bib:ha} provides further examples, and \cite{bib:bmv} more still. \section{Power means} The power means constitute a simultaneous generalization of the arithmetic and geometric means; the basic inequality governing them leads to a raft of new statements, and exposes a symmetry in the AM-GM inequality that would not otherwise be evident. For any real number $r \neq 0$, and any positive reals $x_{1}, \dots, x_{n}$, we define the $r$-th power mean of the $x_{i}$ as \[ M^{r}(x_{1}, \dots, x_{n}) = \left( \frac{x_{1}^{r} + \cdots + x_{n}^{r}}{n} \right)^{1/r}. \] More generally, if $w_{1}, \dots, w_{n}$ are positive real numbers adding up to 1, we may define the weighted $r$-th power mean \[ M^{r}_{w}(x_{1}, \dots, x_{n}) = \left( w_{1} x_{1}^{r} + \cdots + w_{n} x_{n}^{r} \right)^{1/r}. \] Clearly this quantity is continuous as a function of $r$ (keeping the $x_{i}$ fixed), so it makes sense to define $M^{0}$ as \begin{align*} \lim_{r \to 0} M^{r}_{w} &= \lim_{r \to 0} \left( \frac{1}{r} \exp \log (w_{1} x_{1}^{r} + \cdots + w_{n} x_{n}^{r} \right) \\ &= \exp \left. \frac{d}{dr} \right|_{r=0} \log (w_{1} x_{1}^{r} + \cdots + w_{n} x_{n}^{r} ) \\ &= \exp \frac{w_{1} \log x_{1} + \cdots + w_{n} \log x_{n}}{w_{1} + \cdots + w_{n}} = x_{1}^{w_{1}} \cdots x_{n}^{w_{n}} \end{align*} or none other than the weighted geometric mean of the $x_{i}$. \begin{theorem}[Power mean inequality] If $r > s$, then \[ M^{r}_{w}(x_{1}, \dots, x_{n}) \geq M^{s}_{w}(x_{1}, \dots, x_{n}) \] with equality if and only if $x_{1} = \cdots = x_{n}$. \end{theorem} \begin{proof} Everything will follow from the convexity of the function $f(x) = x^{r}$ for $r \geq 1$ (its second derivative is $r(r-1)x^{r-2}$), but we have to be a bit careful with signs. Also, we'll assume neither $r$ nor $s$ is nonzero, as these cases can be deduced by taking limits. First suppose $r >s > 0$. Then Jensen's inequality for the convex function $f(x) = x^{r/s}$ applied to $x_{1}^{s}, \dots, x_{n}^{s}$ gives \[ w_{1} x_{1}^{r} + \cdots + w_{n} x_{n}^{r} \geq \left( w_{1} x_{1}^{s} + \cdots + w_{n} x_{n}^{s} \right)^{r/s} \] and taking the $1/r$-th power of both sides yields the desired inequality. Now suppose $0 > r > s$. Then $f(x) = x^{r/s}$ is concave, so Jensen's inequality is reversed; however, taking $1/r$-th powers reverses the inequality again. Finally, in the case $r > 0 > s$, $f(x) = x^{r/s}$ is again convex, and taking $1/r$-th powers preserves the inequality. (Or this case can be deduced from the previous ones by comparing both power means to the geometric mean.) \end{proof} Several of the power means have specific names. Of course $r=1$ yields the arithmetic mean, and we defined $r=0$ as the geometric mean. The case $r=-1$ is known as the \emph{harmonic mean}, and the case $r=2$ as the \emph{quadratic mean} or \emph{root mean square}. % %RMS-AM-GM-HM. \begin{exer} \ii (Russia, 1995) Prove that for $x,y > 0$, \[ \frac{x}{x^{4} + y^{2}} + \frac{y}{y^{4} + x^{2}} \leq \frac{1}{xy}. \] \ii (Romania, 1996) Let $x_{1}, \dots, x_{n+1}$ be positive real numbers such that $x_{1} + \cdots + x_{n} = x_{n+1}$. Prove that \[ \sum_{i=1}^{n} \sqrt{x_{i}(x_{n+1}-x_{i})} \leq \sqrt{\sum_{i=1}^{n} x_{n+1}(x_{n+1}-x_{i})}. \] \ii (Poland, 1995) For a fixed integer $n \geq 1$ compute the minimum value of the sum \[ x_{1} + \frac{x_{2}^{2}}{2} + \cdots + \frac{x_{n}^{n}}{n} \] given that $x_{1}, \dots, x_{n}$ are positive numbers subject to the condition \[ \frac{1}{x_{1}} + \cdots + \frac{1}{x_{n}} = n. \] \ii Let $a,b,c,d\geq0$. Prove that \[\frac 1a+\frac 1b+\frac 4c + \frac{16}{d}\geq\frac{64}{a+b+c+d}.\] \ii Extend the Rado/Popoviciu inequalities to power means by proving that \[ \frac{M^{r}_{w}(x_{1}, \dots, x_{n})^{k} - M^{s}_{w}(x_{1}, \dots, x_{n})^{k}}{w_n} \geq \frac{M^{r}_{w'}(x_{1}, \dots, x_{n-1})^{k} - M^{s}_{w'}(x_{1}, \dots, x_{n-1})^{k}}{w'_{n-1}} \] whenever $r \geq k \geq s$. Here the weight vector $w' = (w'_1, \dots, w'_{n-1})$ is given by $w'_i = w_i/(w_1 + \cdots + w_{n-1})$. (Hint: the general result can be easily deduced from the cases $k=r,s$.) \ii Prove the following ``converse'' to the power mean inequality: if $r > s> 0$, then \[\Lp\sum_i x_i^r\Rp^{1/r} \leq\Lp\sum_i x_i^s\Rp^{1/s}.\] %\ii %Give another proof of the power mean inequality by showing directly %that the derivative in $r$ of $M_w^r(x_1,\dots, x_n)$ is nonnegative. \end{exer} \section{Cauchy-Schwarz, H\"older and Minkowski inequalities} This section consists of three progressively stronger inequalities, beginning with the simple but versatile Cauchy-Schwarz inequality. \begin{theorem}[Cauchy-Schwarz] For any real numbers $a_{1}, \dots, a_{n}$, $b_{1}, \dots, b_{n}$, \[ (a_{1}^{2} + \cdots + a_{n}^{2})(b_{1}^{2} + \cdots + b_{n}^{2}) \geq (a_{1}b_{1} + \cdots + a_{n}b_{n})^{2}, \] with equality if the two sequences are proportional. \end{theorem} \begin{proof} The difference between the two sides is \[ \sum_{i s$ be nonzero real numbers. Then for any positive real numbers $a_{ij}$, \[ \left( \sum_{j=1}^{m} \left( \sum_{i=1}^{n} a_{ij}^{r} \right)^{s/r} \right)^{1/s} \geq \left( \sum_{i=1}^{n} \left( \sum_{j=1}^{m} a_{ij}^{s} \right)^{r/s} \right)^{1/r}. \] \end{theorem} Minkowski's theorem may be interpreted as a comparison between taking the $r$-th power mean of each row of a matrix, then taking the $s$-th power mean of the results, versus taking the $s$-th power mean of each column, then taking the $r$-th power mean of the result. If $r>s$, the former result is larger, which should not be surprising since there we only use the smaller $s$-th power mean operation once. This interpretation also tells us what Minkowski's theorem should say in case one of $r,s$ is zero. The result is none other than H\"older's inequality! \begin{proof} First suppose $r>s>0$. Write $L$ and $R$ for the left and right sides, respectively, and for convenience, write $t_i = \sum_{j=1}^m a_{ij}^s$. Then \begin{eqnarray*} R^r &=& \sum_{i=1}^n t_i^{r/s} = \sum_{i=1}^n t_i t_i^{r/s - 1} \\ &=& \sum_{j=1}^m \left( \sum_{i=1}^n a_{ij}^s t_i^{r/s - 1} \right) \\ &\leq& \sum_{j=1}^m \left( \sum_{i=1}^n a_{ij}^r \right)^{s/r} \left( \sum_{i=1}^n t_i^{r/s} \right)^{1-s/r} = L^s R^{r-s}, \end{eqnarray*} where the one inequality is a (term-by-term) invocation of H\"older. Next suppose $r>0>s$. Then the above proof carries through provided we can prove a certain variation of H\"older's inequality with the direction reversed. We have left this to you (Problem~\ref{backholder}). Finally suppose $0>r>s$. Then replacing $a_{ij}$ by $1/a_{ij}$ for all $i,j$ turns the desired inequality into another instance of Minkowski, with $r$ and $s$ replaced by $-s$ and $-r$. This instance follows from what we proved above. \end{proof} \begin{exer} \ii (Iran, 1998) Let $x,y,z$ be real numbers not less than 1, such that $1/x + 1/y + 1/z = 2$. Prove that \[ \sqrt{x+y+z} \geq \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1}. \] \ii \label{backholder} Complete the proof of Minkowski's inequality by proving that if $k < 0$ and $a_1, \dots, a_n$, $b_1, \dots, b_n > 0$, then \[ a_1^k b_1^{1-k} + \cdots + a_n^k b_n^{1-k} \geq (a_1 + \cdots + a_n)^k (b_1 + \cdots + b_n)^{1-k}. \] (Hint: reduce to H\"older.) \ii Formulate a weighted version of Minkowski's inequality, with one weight corresponding to each $a_{ij}$. Then show that this apparently stronger result follows from Theorem~\ref{minkowski}! \ii (Titu Andreescu) Let $P$ be a polynomial with positive coefficients. Prove that if \[P\Lp\frac 1x \Rp\geq \frac{1}{P(x)}\] holds for $x=1$, then it holds for every $x>0$. \ii Prove that for $w,x,y,z \geq 0$, \[ w^{6}x^{3} + x^{6} y^{3} + y^{6}z^{3} + z^{6}x^{3} \geq w^{2}x^{5}y^{2} + x^{2}y^{5}z^{2} + y^{2}z^{5}w^{2} + z^{2}x^{5}y^{2}. \] (This can be done either with H\"older or with weighted AM-GM; try it both ways.) \ii (China) Let $r_i, s_i, t_i, u_i, v_i$ be real numbers not less than 1, for $i=1,2,\dots, n$, and let $R,S,T,U,V$ be the respective arithmetic means of the $r_i, s_i, t_i, u_i, v_i$. Prove that \[ \prod_{i=1}^n \frac{r_is_it_iu_iv_i+1}{r_is_it_iu_iv_i-1} \leq \left( \frac{RSTUV+1}{RSTUV-1} \right)^n. \] (Hint: use H\"older and the concavity of $f(x) = (x^5-1)/(x^5+1)$. \ii (IMO 1992/5) Let $S$ be a finite set of points in three-dimensional space. Let $S_x, S_y, S_z$ be the orthogonal projections of $S$ onto the $yz$, $zx$, $xy$ planes, respectively. Show that \[ |S|^2 \leq |S_x| |S_y||S_z|. \] (Hint: Use Cauchy-Schwarz to prove the inequality \[ \left( \sum_{i,j,k} a_{ij} b_{jk} c_{ki} \right)^2 \leq \sum_{i,j} a_{ij}^2 \sum_{j,k} b_{jk}^2 \sum_{k,i} c_{ki}^2. \] Then apply this with each variable set to 0 or 1.) \end{exer} \section{Rearrangement, Chebyshev and ``arrange in order''} Our next theorem is in fact a triviality that every businessman knows: you make more money by selling most of your goods at a high price and a few at a low price than vice versa. Nonetheless it can be useful! (In fact, an equivalent form of this theorem was on the 1975 IMO.) \begin{theorem}[Rearrangement] Given two increasing sequences $x_{1}, \dots, x_{n}$ and $y_{1}, \dots, y_{n}$ of real numbers, the sum \[ \sum_{i=1}^{n} x_{i}y_{\sigma(i)}, \] for $\sigma$ a permutation of $\{1, \dots, n\}$, is maximized when $\sigma$ is the identity permutation and minimized when $\sigma$ is the reversing permutation. \end{theorem} \begin{proof} For $n=2$, the inequality $ac + bd \geq ad + bc$ is equivalent, after collecting terms and factoring, to $(a-b)(c-d) \geq 0$, and so holds when $a \geq b$ and $c \geq d$. We leave it to the reader to deduce the general case from this by successively swapping pairs of the $y_i$. Another way to state this argument involves ``partial summation'' (which the calculus-equipped reader will recognize as analogous to doing integration by parts). Let $s_i = y_{\sigma(i)} + \cdots + y_{\sigma(n)}$ for $i = 1, \dots, n$, and write \[ \sum_{i=1}^n x_i y_{\sigma(i)} = x_1 s_1 + \sum_{j=2}^n (x_j - x_{j-1}) s_j. \] The desired result follows from this expression and the inequalities \[ x_1 + \cdots + x_{n-j+1} \leq s_j \leq x_j + \cdots + x_n, \] in which the left equality holds for $\sigma$ equal to the reversing permutation and the right equality holds for $\sigma$ equal to the identity permutation. \end{proof} One important consequence of rearrangement is Chebyshev's inequality. \begin{theorem}[Chebyshev] Let $x_{1}, \dots, x_{n}$ and $y_{1}, \dots, y_{n}$ be two sequences of real numbers, at least one of which consists entirely of positive numbers. Assume that $x_{1} < \cdots < x_{n}$ and $y_{1} < \cdots < y_{n}$. Then \[ \frac{x_{1} + \cdots + x_{n}}{n} \cdot \frac{y_{1} + \cdots + y_{n}}{n} \leq \frac{x_{1}y_{1} + \cdots + x_{n}y_{n}}{n}. \] If instead we assume $x_{1} > \cdots > x_{n}$ and $y_{1} < \cdots < y_{n}$, the inequality is reversed. \end{theorem} \begin{proof} Apply rearrangement to $x_1, x_2, \dots, x_n$ and $y_1, \dots, y_n$, but repeating each number $n$ times. \end{proof} The great power of Chebyshev's inequality is its ability to split a sum of complicated terms into two simpler sums. We illustrate this ability with yet another solution to IMO 1995/5. Recall that we need to prove \[ \frac{x^{2}}{y+z} + \frac{y^{2}}{z+x} + \frac{z^{2}}{x+y} \geq \frac{x+y+z}{2}. \] Without loss of generality, we may assume $x \leq y \leq z$, in which case $1/(y+z) \leq 1/(z+x) \leq 1/(x+y)$. Thus we may apply Chebyshev to deduce \[ \frac{x^{2}}{y+z} + \frac{y^{2}}{z+x} + \frac{z^{2}}{x+y} \geq \frac{x^{2}+y^{2}+z^{2}}{3} \cdot \left( \frac{1}{y+z} + \frac{1}{z+x} + \frac{1}{x+y} \right). \] By the Power Mean inequality, \begin{align*} \frac{x^{2} + y^{2} + z^{2}}{3} &\geq \left(\frac{x + y + z}{3}\right)^{2} \\ \frac{1}{3} \left(\frac{1}{y+z} + \frac{1}{z+x} + \frac{1}{x+y}\right) \geq \frac{3}{x+y+z} \end{align*} and these three inequalities constitute the proof. The rearrangement inequality, Chebyshev's inequality, and Schur's inequality are all general examples of the ``arrange in order'' principle: when the roles of several variables are symmetric, it may be profitable to desymmetrize by assuming (without loss of generality) that they are sorted in a particular order by size. \begin{exer} \ii Deduce Schur's inequality from the rearrangement inequality. \ii For $a,b,c >0$, prove that $a^a b^b c^c \geq a^b b^c c^a$. \ii (IMO 1971/1) Prove that the following assertion is true for $n=3$ and $n=5$ and false for every other natural number $n>2$: if $a_1, \dots, a_n$ are arbitrary real numbers, then \[ \sum_{i=1}^n \prod_{j \neq i} (a_i - a_j) \geq 0. \] \ii (Proposed for 1999 USAMO) Let $x,y,z$ be real numbers greater than 1. Prove that \[ x^{x^2+2yz} y^{y^2+2zx} z^{z^2+2xy} \geq (xyz)^{xy+yz+zx}. \] \end{exer} \section{Bernoulli's inequality} The following quaint-looking inequality occasionally comes in handy. \begin{theorem}[Bernoulli] For $r >1$ and $x > -1$ (or $r$ an even integer and $x$ any real), $(1 + x)^{r} \geq 1 + xr$. \end{theorem} \begin{proof} The function $f(x) = (1+x)^{r}$ has second derivative $r(r-1)(1+x)^{r-2}$ and so is convex. The claim is simply the fact that this function lies above its tangent line at $x=0$. \end{proof} \begin{exer} \ii (USA, 1992) Let $a = (m^{m+1}+n^{n+1})/(m^m+n^n)$ for $m,n$ positive integers. Prove that $a^m + a^n \geq m^m + n^n$. (The problem came with the hint to consider the ratio $(a^N-N^N)/(a-N)$ for real $a \geq 0$ and integer $N \geq 1$. On the other hand, you can prove the claim for real $m,n \geq 1$ using Bernoulli.) \end{exer} \chapter{Calculus} \label{chap:calc} To some extent, I regard this section, as should you, as analogous to sex education in school--it doesn't condone or encourage you to use calculus on Olympiad inequalities, but rather seeks to ensure that if you insist on using calculus, that you do so properly. Having made that disclaimer, let me turn around and praise calculus as a wonderful discovery that immeasurably improves our understanding of real-valued functions. If you haven't encountered calculus yet in school, this section will be a taste of what awaits you---but no more than that; our treatment is far too compressed to give a comprehensive exposition. After all, isn't that what textbooks are for? Finally, let me advise the reader who thinks he/she knows calculus already not to breeze too this section too quickly. Read the text and make sure you can write down the proofs requested in the exercises. \section{Limits, continuity, and derivatives} The definition of limit attempts to formalize the idea of evaluating a function at a point where it might not be defined, by taking values at points ``infinitely close'' to the given point. Capturing this idea in precise mathematical language turned out to be a tricky business which remained unresolved until long after calculus had proven its utility in the hands of Newton and Leibniz; the definition we use today was given by Weierstrass. Let $f$ be a function defined on an open interval, except possibly at a single point $t$. We say the \emph{limit of $f$ at $t$} exists and equals $L$ if for each positive real $\epsilon$, there exists a positive real $\delta$ (depending on $\epsilon$, of course) such that \[ 0< |x - t| < \delta \implies |f(x) - L| <\epsilon. \] We notate this $\lim_{x\to t}f(x) = L$. If $f$ is also defined at $t$ and $f(t) = L$, we say $f$ is \emph{continuous} at $t$. Most common functions (polynomials, sine, cosine, exponential, logarithm) are continuous for all real numbers (the logarithm only for positive reals), and those which are not (rational functions, tangent) are continuous because they go to infinity at certain points, so the limits in question do not exist. The first example of an ``interesting'' limit occurs in the definition of the derivative. Again, let $f$ be a function defined on an open interval and $t$ a point in that interval; this time we require that $f(t)$ is defined as well. If the limit \[ \lim_{x \to t} \frac{f(x)-f(t)}{x-t} \] exists, we call that the \emph{derivative} of $f$ at $t$ and notate it $f'(t)$. We also say that $f$ is \emph{differentiable} at $t$ (the process of taking the derivative is called \emph{differentiation}). For example, if $f(x) = x^2$, then for $x \neq t$, $(f(x)-f(t))/(x-t) = x+t$, and so the limit exists and $f'(t) = 2t$. More generally, you will show in the exercises that for $f(x) = x^n$, $f'(t) = nt^{n-1}$. On the other hand, for other functions, e.g.\ $f(x) = \sin x$, the difference quotient cannot be expressed in closed form, so certain inequalities must be established in order to calculate the derivative (as will also take place in the exercises). The derivative can be geometrically interpreted as the slope of the tangent line to the graph of $f$ at the point $(t, f(t))$. This is because the tangent line to a curve can be regarded as the limit of secant lines, and the quantity $(f(x+t)-f(x))/t$ is precisely the slope of the secant line through $(x,f(x))$ and $(x+t, f(x+t))$. (The first half of this sentence is entirely unrigorous, but it can be rehabilitated.) It is easy to see that a differentiable function must be continuous, but the converse is not true; the function $f(x) = |x|$ (absolute value) is continuous but not differentiable at $x=0$. An important property of continuous functions, though one we will not have direct need for here, is the intermediate value theorem. This theorem says the values of a continuous function cannot ``jump'', as the tangent function does near $\pi/2$. \begin{theorem} Suppose the function $f$ is continuous on the closed interval $[a,b]$. Then for any $c$ between $f(a)$ and $f(b)$, there exists $x \in [a,b]$ such that $f(x) = c$. \end{theorem} \begin{proof} Any proof of this theorem tends to be a bit unsatisfying, because it ultimately boils down to one's definition of the real numbers. For example, one consequence of the standard definition is that every set of real numbers which is bounded above has a least upper bound. In particular, if $f(a) \leq c$ and $f(b) \geq c$, then the set of $x \in [a,b]$ such that $f(x) < c$ has a least upper bound $y$. By continuity, $f(x) \leq c$; on the other hand, if $f(x) > c$, then $f(x-t) > 0$ for all $t$ less than some positive $\epsilon$, again by continuity, in which case $x-\epsilon$ is an upper bound smaller than $x$, a contradiction. Hence $f(x) = c$. \end{proof} Another important property of continuous functions is the ``extreme value theorem''. \begin{theorem} A continuous function on a closed interval has a global maximum and minimum. \end{theorem} \begin{proof} \end{proof} This statement is of course false for an open or infinite interval; the function may go to infinity at one end, or may approach an extreme value without achieving it. (In technical terms, a closed interval is \emph{compact} while an open interval is not.) \begin{exer} \ii Prove that $f(x) = x$ is continuous. \ii Prove that the product of two continuous functions is continuous, and that the reciprocal of a continuous function is continuous at all points where the original function is nonzero. Deduce that all polynomials and rational functions are continuous. \ii (Product rule) Prove that $(fg)' = fg' + f'g$ and find a similar formula for $(f/g)'$. (No, it's not $f'/g - f/g'$. Try again.) \ii (Chain rule) If $h(x) = f(g(x))$, prove that $h'(x) = f'(g(x)) g'(x)$. \ii Show that the derivative of $x^{n}$ is $n x^{n-1}$ for $n \in \ZZ$. \ii Prove that $\sin x < x < \tan x$ for $0 < x < \pi/2$. (Hint: use a geometric interpretation, and remember that $x$ represents an angle in radians!) Conclude that $\lim_{x \to 0} \sin x / x = 1$. \ii Show that the derivative of $\sin x$ is $\cos x$ and that of $\cos x$ is $(-\sin x)$. While you're at it, compute the derivatives of the other trigonometric functions. \ii Show that an increasing function on a closed interval satisfying the intermediate value theorem must be continuous. (Of course, this can fail if the function is not increasing!) In particular, the functions $c^{x}$ (where $c>0$ is a constant) and $\log x$ are continuous. \ii For $c > 0$ a constant, the function $f(x) = (c^{x} - 1)/x$ is continuous for $x \neq 0$ by the previous exercise. Prove that its limit at 0 exists. (Hint: if $c>1$, show that $f$ is increasing for $x \neq 0$; it suffices to prove this for rational $x$ by continuity.) \ii Prove that the derivative of $c^{x}$ equals $k c^{x}$ for some constant $k$. (Use the previous exercise.) Also show (using the chain rule) that $k$ is proportional to the logarithm of $c$. In fact, the base $e$ of the natural logarithms is defined by the property that $k=1$ when $c=e$. \ii Use the chain rule and the previous exercise to prove that the derivative of $\log x$ is $1/x$. (Namely, take the derivative of $e^{\log x} = x$.) \ii (Generalized product rule) Let $f(y,z)$ be a function of two variables, and let $g(x) = f(x,x)$. Prove that $g'(x)$ can be written as the sum of the derivative of $f$ as a function of $y$ alone (treating $z$ as a constant function) and the derivative of $f$ as a function of $z$ alone, both evaluated at $y=z=x$. For example, the derivative of $x^x$ is $x^x \log x + x^x$. \end{exer} \section{Extrema and the first derivative} The derivative can be used to detect local extrema. A point $t$ is a \emph{local maximum} (resp. minimum) for a function $f$ if $f(t) \geq f(x)$ (resp. $f(t) \leq f(x)$) for all $x$ in some open interval containing $t$. \begin{theorem} If $t$ is a local extremum for $f$ and $f$ is differentiable at $t$, then $f'(t) = 0$. \end{theorem} \begin{cor}[Rolle] If $f$ is differentiable on the interval $[a,b]$ and $f(a)=f(b)=0$, then there exists $x\in [a,b]$ such that $f'(x) = 0$. \end{cor} So for example, to find the extrema of a continuous function on a closed interval, it suffices to evaluate it at \begin{itemize} \ii all points where the derivative vanishes, \ii all points where the derivative is not defined, and \ii the endpoints of the interval, \end{itemize} since we know the function has global minima and maxima, and each of these must occur at one of the aforementioned points. If the interval is open or infinite at either end, one must also check the limiting behavior of the function there. A special case of what we have just said is independently useful: if a function is positive at the left end of an interval and has nonnegative derivative on the interval, it is positive on the entire interval. \section{Convexity and the second derivative} As noted earlier, a twice differentiable function is convex if and only if its second derivative is nonnegative. \section{More than one variable} \textbf{Warning!} The remainder of this chapter is somewhat rougher going than what came before, in part because we need to start using the language and ideas of linear algebra. Rest assured that none of the following material is referenced anywhere else in the notes. We have a pretty good understanding now of the relationship between extremization and the derivative, for functions in a single variable. However, most inequalities in nature deal with functions in more than one variable, so it behooves us to extend the formalism of calculus to this setting. Note that whatever we allow the domain of a function to be, the \emph{range} of our ``functions'' will always be the real numbers. (We have no need to develop the whole of multivariable calculus when all of our functions arise from extremization questions!) The formalism becomes easier to set up in the language of linear algebra. If $f$ is a function from a vector space to $\RR$ defined in a neighborhood of (i.e.\ a ball around) a point $\vx$, then we say the limit of $f$ at $\vx$ exists and equals $L$ if for every $\epsilon > 0$, there exists $\delta > 0$ such that \[ 0 < ||\vt|| < \delta \implies |f(\vx + \vt) - L| < \epsilon. \] (The double bars mean length in the Euclidean sense, but any reasonable measure of the length of a vector will give an equivalent criterion; for example, measuring a vector by the maximum absolute value of its components, or by the sum of the absolute values of its components.) We say $f$ is continuous at $\vx$ if $\lim_{\vt \to 0} f(\vx + \vt) = f(\vx)$. If $\vy$ is any vector and $\vx$ is in the domain of $f$, we say the \emph{directional derivative} of $f$ along $\vx$ in the direction $\vy$ exists and equals $f_{\vy}(\vx)$ if \[ f_{\vy}(\vx) = \lim_{t \to 0} \frac{f(\vx + t\vy) - f(\vx)}{t}. \] If $f$ is written as a function of variables $x_1, \dots, x_n$, we call the directional derivative along the $i$-th standard basis vector the \emph{partial derivative} of $f$ with respect to $i$ and denote it by $\frac{\partial f}{\partial x_i}$. In other words, the partial derivative is the derivative of $f$ as a function of $x_i$ along, regarding the other variables as constants. TOTAL DERIVATIVE Caveat! Since the derivative is not a ``function'' in our restricted sense (it has takes values in a vector space, not $\RR$) we cannot take a ``second derivative''---yet. ASSUMING the derivative exists, it can be computed by taking partial derivatives along a basis. \section{Convexity in several variables} A function $f$ defined on a convex subset of a vector space is said to be \emph{convex} if for all $\vx, \vy$ in the domain and $t \in [0,1]$, \[ tf(\vx) + (1-t)f(\vy) \geq f(t\vx + (1-t)\vy). \] Equivalently, $f$ is convex if its restriction to any line is convex. Of course, we say $f$ is \emph{concave} if $-f$ is convex. The analogue of the second derivative test for convexity is the Hessian criterion. A symmetric matrix $M$ (that is, one with $M_{ij} = M_{ji}$ for all $i,j$) is said to be \emph{positive definite} if $M\vx \cdot \vx > 0$ for all nonzero vectors $\vx$, or equivalently, if its eigenvalues are all real and positive. (The first condition implies the second because all eigenvalues of a symmetric matrix are real. The second implies the first because if $M$ has positive eigenvalues, then it has a square root $N$ which is also symmetric, and $M\vx \cdot \vx = (N\vx) \cdot (N\vx) > 0$.) \begin{theorem}[Hessian test] A twice differentiable function $f(x_1, \dots, x_n)$ is convex in a region if and only if the Hessian matrix \[ H_{ij} = \frac{\partial^2}{\partial x_i \partial x_j} \] is positive definite everywhere in the region. \end{theorem} Note that the Hessian is symmetric because of the symmetry of mixed partials, so this statement makes sense. \begin{proof} The function $f$ is convex if and only if its restriction to each line is convex, and the second derivative along a line through $\vx$ in the direction of $\vy$ is (up to a scale factor) just $H \vy \cdot \vy$ evaluated at $\vx$. So $f$ is convex if and only if $H \vy \cdot \vy >0$ for all nonzero $\vy$, that is, if $H$ is positive definite. \end{proof} The bad news about this criterion is that determining whether a matrix is positive definite is not \emph{a priori} an easy task: one cannot check $M\vx \cdot \vx \geq 0$ for every vector, so it seems one must compute all of the eigenvalues of $M$, which can be quite a headache. The good news is that there is a very nice criterion for positive definiteness of a symmetric matrix, due to Sylvester, that saves a lot of work. \begin{theorem}[Sylvester's criterion] An $n \times n$ symmetric matrix of real numbers is positive definite if and only if the determinant of the upper left $k \times k$ submatrix is positive for $k = 1, \dots, n$. \end{theorem} \begin{proof} By the $M \vx \cdot \vx$ definition, the upper left $k \times k$ submatrix of a positive definite matrix is positive definite, and by the eigenvalue definition, a positive definite matrix has positive determinant. Hence Sylvester's criterion is indeed necessary for positive definiteness. We show the criterion is also sufficient by induction on $n$. BLAH. \end{proof} \begin{exer} \ii (IMO 1968/2) Prove that for all real numbers $x_{1}, x_{2}, y_{1}, y_{2}, z_{1}, z_{2}$ with $x_{1}, x_{2} > 0$ and $x_{1}y_{1} > z_{1}^{2}$, $x_{2}y_{2} > z_2$, the inequality \[ \frac{8}{(x_1+x_2)(y_1+y_2) - (z_1+z_2)^2} \leq \frac{1}{x_1y_1-z_1^2} + \frac{1}{x_2y_2-z_2^2} \] is satisfied, and determine when equality holds. (Yes, you really can apply the material of this section to the IMO! Verify convexity of the appropriate function using the Hessian and Sylvester's criterion.) \end{exer} \section{Constrained extrema and Lagrange multipliers} In the multivariable realm, a new phenomenon emerges that we did not have to consider in the one-dimensional case: sometimes we are asked to prove an inequality in the case where the variables satisfy some constraint. The Lagrange multiplier criterion for an interior local extremum of the function $f(x_1, \dots, x_n)$ under the constraint $g(x_1, \dots, x_n) = c$ is the existence of $\lambda$ such that \[ \frac{\partial f}{\partial x_i}(x_1, \dots, x_n) = \lambda \frac{\partial g}{\partial x_i}(x_1, \dots, x_n). \] Putting these conditions together with the constraint on $g$, one may be able to solve and thus put restrictions on the locations of the extrema. (Notice that the duality of constrained optimization shows up in the symmetry between $f$ and $g$ in the criterion.) It is even more critical here than in the one-variable case that the Lagrange multiplier condition is a necessary one only for an \emph{interior} extremum. Unless one can prove that the given function is convex, and thus that an interior extremum must be a global one, one must also check all boundary situations, which is far from easy to do when (as often happens) these extend to infinity in some direction. For a simple example, let $f(x,y,z) = ax + by + cz$ with $a,b,c$ constants, not all zero, and consider the constraint $g(x,y,z)=1$, where $g(x,y,z)= x^2+y^2+z^2$. Then the Lagrange multiplier condition is that \[ a = 2 \lambda x, b = 2 \lambda y, c = 2 \lambda z. \] The only points satisfying this condition plus the original constraint are \[ \pm \frac{1}{\sqrt{a^2+b^2+c^2}} (a,b,c), \] and these are indeed the minimum and maximum for $f$ under the constraint, as you may verify geometrically. %\chapter{Geometric inequalities} \chapter{Coda} \section{Quick reference} Here's a handy reference guide to the techniques we've introduced. \begin{itemize} \ii Arithmetic-geometric-harmonic means \ii Arrange in order \ii Bernoulli \ii Bunching \ii Cauchy-Schwarz \ii Chebyshev \ii Convexity \ii Derivative test \ii Duality for constrained optimization \ii Equality conditions \ii Factoring \ii Geometric interpretations \ii Hessian test \ii H\"older \ii Jensen \ii Lagrange multipliers \ii Maclaurin \ii Minkowski \ii Newton \ii Power means \ii Rearrangement \ii Reduction of the number of variables (Theorem~\ref{trick}) \ii Schur \ii Smoothing \ii Substitution (algebraic or trigonometric) \ii Sum of squares \ii Symmetric sums \ii Unsmoothing (boundary extrema) \end{itemize} \section{Additional problems} Here is an additional collection of problems covering the entire range of techniques we have introduced, and one or two that you'll have to discover for yourselves! \begin{exer} \ii Let $x,y,z>0$ with $xyz=1$. Prove that $x+y+z\leq x^2+y^2+z^2$. \ii The real numbers $x_1,x_2,\ldots, x_n$ belong to the interval $[-1,1]$ and the sum of their cubes is zero. Prove that their sum does not exceed $n/3$. \ii (IMO 1972/2) Let $x_{1}, \dots, x_{5}$ be positive reals such that \[ (x_{i+1}^{2}-x_{i+3}x_{i+5})(x_{i+2}^{2}-x_{i+3}x_{i+5}) \leq 0 \] for $i=1,\dots, 5$, where $x_{n+5}=x_{n}$ for all $n$. Prove that $x_{1} = \cdots = x_{5}$. \ii (USAMO 1979/3) Let $x,y,z\geq 0$ with $x+y+z=1$. Prove that \[ x^3+y^3+z^3+6xyz\geq \frac 14. \] \ii (Taiwan, 1995) Let $P(x) = 1+a_1x+ \cdots + a_{n-1}x^{n-1}+x^n$ be a polynomial with complex coefficients. Suppose the roots of $P(x)$ are $\alf_1,\alf_2, \ldots,\alf_n$ with \[|\alf_1|>1, |\alf_2|>1, \ldots,|\alf_j|>1\] and \[|\alf_{j+1}|\leq1, |\alf_{j+2}|\leq1, \ldots,|\alf_n|\leq1.\] Prove that \[\prod_{i=1}^j |\alf_i| \leq \sqrt{|a_0|^2+|a_1|^2 + \cdots + |a_n|^2}.\] (Hint: look at the coefficients of $P(x) \overline{P}(x)$, the latter being $1 + \overline{a_1} x + \cdots + \overline{a_{n-1}}x^{n-1} + x^n$.) \ii Prove that, for any real numbers $x,y,z$, \[3(x^2-x+1)(y^2-y+1)(z^2-z+1)\geq (xyz)^2+xyz+1.\] \ii \be \ii[(a)] Prove that any polynomial $P(x)$ such that $P(x) \geq 0$ for all real $x$ can be written as the sum of the squares of two polynomials. \ii[(b)] Prove that the polynomial \[ x^2(x^2-y^2)(x^2-1) + y^2(y^2-1)(y^2-x^2) + (1-x^2)(1-y^2) \] is everywhere nonnegative, but cannot be written as the sum of squares of any number of polynomials. (One of Hilbert's problems, solved by Artin and Schreier, was to prove that such a polynomial can always be written as the sum of squares of rational functions.) \ee \end{exer} \begin{thebibliography}{99} \bibitem{bib:bmv} P.S. Bullen, D. Mitrinovi\'c, and P.M.Vasi\'c, \textit{Means and their Inequalities}, Reidel, Dordrecht, 1988. \bibitem{bib:ha} G. Hardy, J.E. Littlewood, and G. P\'olya, \textit{Inequalities} (second edition), Cambridge University Press, Cambridge, 1951. \bibitem{bib:pop} T. Popoviciu, Asupra mediilor aritmetice \,si medie geometrice (Romanian), \textit{Gaz. Mat. Bucharest} \textbf{40} (1934), 155-160. \bibitem{bib:rab} S. Rabinowitz, \textit{Index to Mathematical Problems 1980-1984}, Mathpro Press, Westford (Massachusetts), 1992. \end{thebibliography} \end{document}