# Proofs of AM-GM inequality

The arithmetic - geometric mean inequality states that $$\frac{x_1+ \ldots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}$$ I'm looking for some original proofs of this inequality. I can find the usual proofs on the internet but I was wondering if someone knew a proof that is unexpected in some way. e.g. can you link the theorem to some famous theorem, can you find a non-trivial geometric proof (I can find some of those), proofs that use theory that doesn't link to this inequality at first sight (e.g. differential equations …)?

Induction, backward induction, use of Jensen inequality, swapping terms, use of Lagrange multiplier, proof using thermodynamics (yeah, I know, it's rather some physical argument that this theorem might be true, not really a proof), convexity, … are some of the proofs I know.

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What have you seen, so that we don't repeat what you already know? –  chubakueno Feb 26 at 21:45
AoPS has some interesting ones: artofproblemsolving.com/Wiki/index.php/Proofs_of_AM-GM –  Hayden Feb 26 at 21:46
Would you include that in your post? (And,thermodynamics? ._. ) –  chubakueno Feb 26 at 21:54
I'm curious about the thermodynamics proof $\ddot \smile$ –  dani_s Feb 26 at 22:08
Community wiki? –  nayrb Feb 26 at 22:28

Pólya's Proof:

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

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

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Almost all your inequalities are reversed. –  Martín-Blas Pérez Pinilla Feb 26 at 23:18
@Martín-BlasPérezPinilla Woops, fixed –  qwr Feb 26 at 23:23

As requested by dani_s, I will give the thermodynamic proof of the AM-GM inequality. This is certainly an example of an original proof, although you might argue about whether or not it's rigorous.

Let's start with a list of numbers $x_i$ for which we want to prove the inequality. Take $n$ identical heat reservoirs with the same heat capacity $c$. Reservoir $i$ had initial temperature $x_i$. Bring those reservoirs in contact with each other such that this system evolves to an equilibrium temperature A.

The first law of thermodynamics (conservation of energy) implies that A equals the arithmetic mean of the $x_i$, AM.

The second law of thermodynamics states that the entropy increases until the equilibrium is reached, where the entropy has a maximum. The corresponding formula of change in entropy is: $$\Delta S=c \ln{\frac{T}{T_0}}$$ $c$ is the heat capacity, $T_0$ the initial temperature and $T$ the end temperature.

In our case $T_i=A$ for all $i$ and $T_{0,i}=x_i$. The total entropy didn't decrease and therefore, $$\sum_{i=1}^n c \ln\frac{A}{x_i} \geq 0$$

By writing the sum of logarithms as a logarithm of a product, we recognize the geometric mean. Therefore (since $A=AM$): $$\frac{AM^n}{GM^n} \geq 1$$ This prooves the AM-GM inequality.

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It follows from Jensen's inequality that $$\sqrt{x_1 \cdot x_2} \leq \frac{x_1+x_2}{2}.$$ Applying this inequality twice, we get $$(x_1 x_2 x_3 x_4)^{\frac{1}{4}} \leq \frac{\sqrt{x_1 x_2}+\sqrt{x_3 x_4}}{2} \leq \frac{x_1+x_2+x_3+x_4}{4}.$$ By induction, it is not difficult to see that $$(x_1 \cdots x_{2^k})^{\frac{1}{2^k}} \leq \frac{x_1+\ldots+x_{2^k}}{2^k} \tag{1}$$ for all $k \geq 1$.

It remains to fill the gaps between the powers of two. So let $x_1,\ldots,x_n$ be arbitrary positive numbers and choose $k$ such that $n\leq 2^k$. We set

$$\alpha_i := \begin{cases} x_i & i \leq n \\ A & n< i \leq 2^k \end{cases}$$

where $A:= \frac{x_1+\ldots+x_n}{n}$. Applying $(1)$ to the $(\alpha_1,\ldots,\alpha_{2^k})$ yields

$$\bigg( x_1 \ldots x_n A^{2^k-n} \bigg)^{\frac{1}{2^k}} \leq \frac{x_1+\ldots+x_n+(2^k-n) A}{2^k} = A.$$

Hence,

$$(x_1 \ldots x_n)^{1/n} \leq A = \frac{x_1+\ldots+x_n}{n}.$$

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Bernoulli's Inequality says that for $u\ge-1$ and $0\le r\le1$, $$(1+u)^r\le1+ru\tag{1}$$ Setting $u=\frac xy-1$ in $(1)$ says that for $x,y\gt0$, $$\left(\frac xy\right)^r\le(1-r)+r\frac xy\tag{2}$$ If we multiply $(2)$ by $y$, we get $$x^ry^{1-r}\le rx+(1-r)y\tag{3}$$ Now $(3)$ can be used inductively to get $$x_1^{r_1}x_2^{r_2}x_3^{r_3}\dots x_n^{r_n}\le r_1x_1+r_2x_2+r_3x_3+\dots+r_nx_n\tag{4}$$ where $r_1,r_2,r_3,\dots,r_n\ge0$ and $r_1+r_2+r_3+\dots+r_n=1$.
Suppose that $(4)$ holds, then we can use $(3)$ to get \begin{align} &\left(x_1^{r_1}x_2^{r_2}x_3^{r_3}\dots x_n^{r_n}\right)^{1-r_{n+1}}x_{n+1}^{r_{n+1}}\\ &\le(1-r_{n+1})\left(r_1x_1+r_2x_2+r_3x_3+\dots+r_nx_n\right)+r_{n+1}x_{n+1}\tag{5} \end{align} where $(1-r_{n+1})(r_1+r_2+r_3+\dots+r_n)+r_{n+1}=1$