Proofs of AM-GM inequality

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This is a fairly old answer of mine with a proof that was not very motivated, so I thought I’d update it with some geometric intuition about convexity, which is a good way to understand some inequalities (including Hölder’s Inequality, with Cauchy–Schwarz Inequality as a special case).

Consider for simplicity the two-variable case $(x+y)/2 \ge \sqrt{xy}$:

I’m not sure if it comes across in the diagram, but the arithmetic mean will always produce a flat linear plane while the geometric mean will always produce this concave squareroot-like curvy surface which curves downward. Equality is only achieved precisely when $x = y$ and similarly in higher dimensions. The “curve downward” part shows intuitively why the inequality is true for all other values.

Here’s a slice view with fixed $y=1$:

Obviously a plot is not a proof but it does give a useful visual.

The proof for more than two variables presented requires elementary properties of logarithms, which uses a common trick using log to transform dealing with multiplication to dealing with addition, which is linear. Starting off with the original statement:

$$ \frac{x_1 + \dots + x_n}{n}\ge (x_1 \dots x_n)^{1/n} $$

Taking logs preserves the inequality since $\log$ is an increasing function:

$$\iff \log \left(\frac{x_1 + \dots + x_n}{n}\right) \ge \frac 1 n \log (x_1 \dots x_n) = \frac{\log x_1 + \dots + \log x_n}{n}$$

$\DeclareMathOperator{\E}{E}$
If we write $\E[X]$ as the mean of $x_i$’s and $\E[\log(X)]$ as the mean of $\log x_i$’s, we can also understand this in the language of expectation:

$$\log(\E[X]) \ge \E[\log (X)]$$

Using the concavity of $\log$, by Jensen’s Inequality (proved inductively starting from the definition of convexity, going back to the linearity of expectation, which ultimately comes from addition), the inequality holds.

Original post of Pólya’s Proof, using similar ideas of convexity of $e^x$:

Let $f(x) = e^{x-1}-x$. The first derivative is $f'(x)=e^{x-1}-1$ and the second derivative is $f”(x) = e^{x-1}$.

$f$ is convex everywhere because $f”(x) > 0$, and has a minimum at $x=1$. Therefore $x \le e^{x-1}$ for all $x$, and the equation is only equal when $x=1$.

Using this inequality we get

$$\frac{x_1}{a} \frac{x_2}{a} \cdots \frac{x_n}{a} \le e^{\frac{x_1}{a}-1} e^{\frac{x_2}{a}-1} \cdots e^{\frac{x_n}{a}-1}$$

with $a$ being the arithmetic mean. The right side simplifies

$$\exp \left(\frac{x_1}{a} -1 \ +\frac{x_1}{a} -1 \ + \cdots + \frac{x_n}{a} -1 \right)$$

$$=\exp \left(\frac{x_1 + x_2 + \cdots + x_n}{a} – n \right) = \exp(n – n) = e^0 = 1$$

Going back to the first inequality

$$\frac{x_1x_2\cdots x_n}{a^n} \le 1$$

So we end with

$$\sqrt[n]{x_1x_2\cdots x_n} \le a$$