Proofs of AM-GM

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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 trictly 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. $\blacksquare$

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. $\blacksquare$