Two Elegant Proofs of the AM-GM Inequality

均值不等式的两个巧妙证明 苏剑林 · 2012-09-26 · Mathematics inequality

I recall that a few years ago, BoJone offered a proof of the arithmetic-mean–geometric-mean inequality, but the proof was a bit long and somewhat dazzling to follow (even though the underlying idea was actually quite simple). Yesterday, in my Mathematical Analysis class, the instructor covered this inequality and gave his own proof by mathematical induction, which still didn't feel as clean or elegant. But it reminded me of two clever proofs I had studied before. However, no matter how hard I racked my brain all day yesterday, I couldn't recall either of them. It was only today that they finally came back to me, so I'm sharing them here, partly for others' benefit and partly as a note to my future self.

For a collection of nonnegative numbers $x_i$, we have

$$\frac{x_1+x_2+...+x_n}{n} \geq \sqrt[n]{x_1 x_2 ... x_n}$$

which we denote as $A_n \geq G_n$

Proof 1: Mathematical Induction

This method isn't exactly simple, but it is remarkably clever—the step of going from n to n+1 is the kind of thing that makes you want to applaud. The proof of Jensen's inequality by induction follows the same recursive idea, and the AM-GM inequality is really just a special case of Jensen's inequality.

Suppose $A_n \geq G_n$ holds; we want to prove $A_{n+1} \geq G_{n+1}$. We have

$$\begin{aligned}&2n A_{n+1}=(n+1)A_{n+1}+(n-1)A_{n+1} \\ =&[x_1 + x_2 +...+x_n]+[x_{n+1}+(n-1)A_{n+1}] \\ \geq &nG_n+n(x_{n+1}\cdot A_{n+1}^{n-1})^{\frac{1}{n}} \\ \geq &2n(G_{n+1}^{n+1}\cdot A_{n+1}^{n-1})^{\frac{1}{2n}}\end{aligned}$$more

Simplifying gives us: $A_{n+1} \geq G_{n+1}$

This completes the induction step from n to n+1. The remaining details are omitted.

Proof 2: The Logarithmic Method

This method is even more clever and much simpler—quite possibly the simplest proof I've ever come across. It relies on a very simple formula: for all nonnegative x, $e^x \geq 1+x$, and when $x\geq -1$, both sides are nonnegative.

(For typographical convenience, let $exp(x)=e^x$)

$$\begin{aligned}&\exp\left(\frac{n A_n}{G_n}-n\right) \\ =&\exp\left(\frac{x_1}{G_n}-1\right)\cdot \exp\left(\frac{x_2}{G_n}-1\right)\dots\exp\left(\frac{x_n}{G_n}-1\right) \\ \geq &\frac{x_1}{G_n}\cdot \frac{x_2}{G_n}...\frac{x_n}{G_n} \quad\text{(前面的各项指数部分显然都大于等于-1)}\\ =&1\end{aligned}$$

That is to say

$$\exp\left(n\frac{A_n}{G_n}-n\right) \geq 1$$

A slight simplification then gives $A_n \geq G_n$