Proofs of AM-GM

From AoPSWiki

Revision as of 18:34, 15 September 2008 by 1=2 (Talk | contribs)

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

This pages lists some proofs of the weighted AM-GM Inequality. The inequality's statement is as follows: for all nonnegative reals $a_1, \dotsc, a_n$ and nonnegative reals $\lambda_1, \dotsc, \lambda_n$ such that $\sum_{i=1}^n \lambda_i = 1$ , then $\sum_{i=1}^n \lambda_i a_i \ge \prod_{i=1}^n a_i^{\lambda_i},$ with equality if and only if $a_i = a_j$ for all $i,j$ such that $\lambda_i, \lambda_j \neq 0$ .

We first note that we may disregard any $a_i$ for which $\lambda_i= 0$ , as they contribute to neither side of the desired inequality. We also note that if $a_i= 0$ and $\lambda_i \neq 0$ , for some $i$ , then the right-hand side of the inequality is zero and the right hand of the inequality is greater or equal to zero, with equality if and only if $a_j = 0 = a_i$ whenever $\lambda_j\neq 0$ . Thus we may henceforth assume that all $a_i$ and $\lambda_i$ are strictly positive.

Complete Proofs

Proof by Convexity

We note that the function $x \mapsto \ln x$ is strictly concave. Then by Jensen's Inequality, $\ln \sum_i \lambda_i a_i \ge \lambda_i \sum_i \ln a_i = ln \prod_i a_i^{\lambda_i} ,$ with equality if and only if all the $a_i$ are equal. Since $x \mapsto \ln x$ is a strictly increasing function, it then follows that $\sum_i \lambda_i a_i \ge \prod_i a_i^{\lambda_i},$ with equality if and only if all the $a_i$ are equal, as desired. $\blacksquare$

Alternate Proof by Convexity

This proof is due to G. Pólya.

Note that the function $f:x \mapsto e^x$ is strictly convex. Let $g(x)$ be the line tangent to $f$ at $(0,1)$ ; then $g(x) = x+1$ . Since $f$ is also a continuous, differentiable function, it follows that $f(x) \ge g(x)$ for all $x$ , with equality exactly when $x=0$ , i.e., $1+x \le e^x,$ with equality exactly when $x=0$ .

Now, set $r_i = a_i/\biggl( \sum_{j=1}^n \lambda_j a_j \biggr) - 1,$ for all integers $1\le i \le n$ . Our earlier bound tells us that $r_i +1 \le \exp(r_i),$ so $(r_i +1)^{\lambda_i} \le \exp(\lambda_i r_i) .$ Multiplying $n$ such inequalities gives us $\frac{\prod_{i=1}^n a_i^{\lambda_i}}{ \sum_{i=1}^n \lambda_i r_i} = \prod_{i=1}^n (r_i + 1) \le \prod_{i=1}^n \exp \lambda_i ...$ so $\prod_{i=1}^n a_i^{\lambda_i} \le \sum_{i=1}^n \lambda_i r_i ,$ as desired. Equality holds when $r_i$ is always equal to zero, i.e., when $a_i = a_j$ for all $i,j$ such that $\lambda_i , \lambda_j \neq 0$ . $\blacksquare$

Proofs of Unweighted AM-GM

These proofs use the assumption that $\lambda_i = 1/n$ , for all integers $1 \le i \le n$ .

Proof by Rearrangement

Define the $n$ sequence $\{ r_{ij}\}_{i=1}^{n}$ as $r_{ij} = \sqrt[n]{a_i}$ , for all integers $1 \le i,j \le n$ . Evidently these sequences are similarly sorted. Then by the Rearrangement Inequality, $\sum_i a_i = \sum_i \prod_j r_{ij} \ge \sum_i \prod_j r_{i,i+j} = n \prod_i \sqrt[n]{a_i} ,$ where we take our indices modulo $n$ , with equality exactly when all the $r_{ij}$ , and therefore all the $a_i$ , are equal. Dividing both sides by $n$ gives the desired inequality. $\blacksquare$

Proof by Cauchy Induction

We first prove that the inequality holds for two variables. Note that $(\sqrt{a}, \sqrt{b}), (\sqrt{a}, \sqrt{b})$ are similarly sorted sequences. Then by the Rearrangement Inequality, $\sqrt{ab} = \frac{ \sqrt{a} \sqrt{b} + \sqrt{b} \sqrt{a}}{2} \le \frac{ \sqrt{a} \sqrt{a} + \sqrt{b} \sqrt{b}}{2} = \frac{a+b...$ with equality exactly when $a=b$ .

We next prove that if the inequality holds for $n$ variables (with equality when all are equal to zero), than it holds for $2n$ variables (with equality when all are equal to zero). Indeed, suppose the inequality holds for $n$ variables. Let $A_n, A'_n, A_{2n}$ denote the arithmetic means of $(a_1, \dotsc, a_n)$ , $(a_{n+1}, \dotsc, a_{2n})$ , $(a_1, \dotsc, a_{2n})$ , respectively; let $G_n, G'_n, G_{2n}$ denote their respective geometric means. Then $G_{2n} = \sqrt{G_n G'_n} \le \frac{G_n + G'_n}{2} \le \frac{A_n + A'_n}{2} = A_2n,$ with equality when all the numbers $a_1, \dotsc, a_{2n}$ are equal, as desired.

These two results show by induction that the theorem holds for $n=2^k$ , for all integers $k$ . In particular, for every integer $n$ , there is an integer $N >n$ such that the theorem holds for $N$ variables.