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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 '14 at 21:45
Would you include that in your post? (And,thermodynamics? ._. ) – chubakueno Feb 26 '14 at 21:54
I'm curious about the thermodynamics proof $\ddot \smile$ – dani_s Feb 26 '14 at 22:08
Community wiki? – nayrb Feb 26 '14 at 22:28
Please add the proofs you know as answers (just to get them all together). – vonbrand Feb 27 '14 at 1:25

10 Answers 10

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 '14 at 23:18
@Martín-BlasPérezPinilla Woops, fixed – qwr Feb 26 '14 at 23:23
This reminds me of a trick I found: $e^x \geq 1+x$ by similar calculations as the start of this answer. Then $e^{1/n} > 1 + 1/n = (n+1)/n,$ so $e^{1 + 1/2 + 1/3 + \ldots + 1/n} > (2/1) (3/2) \ldots ( (n+1)/n ) = n+1,$ so $H_n:= 1 + 1/2 + \ldots + 1/n > \log (n+1),$ and in particular it diverges. – Ragib Zaman Jun 12 '14 at 4:17

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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There is no need to argue whether this is or not a rigorous proof, as it isn't. – Mariano Suárez-Alvarez Jun 17 '15 at 16:52

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

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


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

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I believe this one is due to Cauchy. – Siméon Jun 11 '14 at 19:00
Applying the general Jensen inequality ($n$ terms, not just two) it gets much simpler, see eg Wikipedia – leonbloy Jun 12 '14 at 14:16

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 $r_1+r_2+r_3+\dots+r_n=1$.

Inductive step:

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 $(1-r_{n+1})(r_1+r_2+r_3+\dots+r_n)+r_{n+1}=1$

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I shall provide a simple geometric proof of the inequality in the case of two variables (which I have not been able to find anywhere else - a proof involving a triangle in a circle seems to be popular).

Consider the square of side $a + b$ in the figure below.

The area of the square is $(a + b)^2$. But as it completely contains the four blue rectangles, each of area $ab$, it follows that

$(a + b)^2 \ge 4ab \Rightarrow\\ \dfrac{a + b}{2} \ge \sqrt{ab} $

Further, note that there is a square in middle, of side $(b - a)$, and hence area $(b - a)^2$. Therefore the inequality is strict except when $a = b$.

This proves the two-variable case. The same can be extended to the $n$-variable case. I have tried extending it to three variables, but it is difficult to argue why exactly $27$ rectangular parallelepipeds (of sides $a, b, c$) fit in the cube (of side $a + b + c$), though I can see it is so. Any suggestions?

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There is a more direct induction proof.

We can assume that $0<a_1\leq \ldots \leq a_n$. Let set $A=\frac{a_1+ \ldots +a_n}{n}$. First notice that $(a_n-A)(A-a_1)\geq 0$, it can be rewritten as $\frac{a_1a_n}{A}\leq a_1+a_2-A $. Now we assume that the property hold for $n-1$. We can apply it with ${a_2,\ldots , a_{n-1},a_1+a_n-A}$. It gives : $$(\frac{\prod_{k=1}^na_k}{A})^{1/(n-1)}\leq [(\prod _{k=2}^{n-1}a_k)(a_1+a_n-A)]^{1/(n-1)}\leq \frac{a_1+\ldots +a_n-A}{n-1}= A.$$

So : $$\prod_{k=1}^na_k\leq (A^{1+1/(n-1)})^{n-1}=A^n.$$

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This is a proof using Buffalo way (see this link for what buffalo way is, I couldn't find it anywhere, I jush heard it's possible, so I tried it, hopefully there are no mistakes, but even if there are I think they will be easy to correct, the main idea should still be right. We'll proceed by strong induction, base case holds trivially from inequality $(x_1 - x_2)^2 \geq 0$, assume that AM-GM holds for all $k$ such that $1 < k < n$. Then we wish to prove that $$x_1^n + x_2^n + \cdots + x_n^n - nx_1x_2\cdots x_n \geq 0$$ This inequality is clearly symmetric, hence we may WLOG suppose $x_1 = \min\{x_1,x_2,\cdots,x_n\} = x$, therefore there exist $y_1,y_2,\cdots,y_{n - 1} \geq 0$ such that $x_i = x + y_{i - 1}$ for all $i \in \{1,2,\cdots,n\}$. Therefore the inequality can be rewritten as $$x^n + (x + y_1)^n + (x + y_2)^n + \cdots + (x + y_{n - 1})^n - nx(x + y_1)(x + y_2) \cdots (x + y_{n - 1}) \geq 0$$ This is a polynomial in $x$ and expanding this polynomial we get that the coefficient of $x^{n - k}$, where $1 < k < n$, is $$\binom{n}{k}p_k(y_1,y_2,\cdots,y_{n - 1}) - ne_k(y_1,y_2,\cdots,y_{n - 1})$$ where $p_k$ is $k$-th power sum and $e_k$ is $k$-th elementary symmetric polynomial. So it suffices to prove that $$\binom{n}{k}p_k(y_1,y_2,\cdots,y_{n - 1}) - ne_k(y_1,y_2,\cdots,y_{n - 1}) \geq 0$$ But by inductive hypothesis we get $$\frac{n}{k} \cdot \left(y^k_{i_1} + y^k_{i_2} + \cdots + y^k_{i_k}\right) - ny_{i_1}y_{i_2} \cdots y_{i_k} \geq 0,$$ where $i_1,i_2,\cdots,i_k \in \{1,2,3,\cdots,n - 1\}$ are pairwise distinct. Doing this for all the possible combinations of indices $i_1,i_2, \cdots, i_k$ we in fact get stronger inequality $$\frac{n}{k} \cdot \binom{n - 2}{k - 1} p_k(y_1,y_2,\cdots,y_{n - 1}) - ne_k(y_1,y_2,\cdots,y_{n - 1}) \geq 0.$$ Now also clearly coefficients of $x^n$ and $x^{n - 1}$ are $0$ and coefficient of $x^0$ is just $p_n(y_1,y_2,\cdots,y_{n - 1})$. Hence all the coefficients of our polynomial are non-negative, therefore the polynomial is non-negative, thus the inductive step is proved and the whole proof is finished.

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Applying the Log sum inequality (direct consequence of Jensen inequality)

$$ \sum a_i \log \frac{ b_i}{a_i} \le a \log \frac{b}{a}$$ where $a=\sum a_i$ and $b=\sum b_i$, $a_i\ge 0$, $b_i\ge 0$; setting $b_i=x_i/n$ $a_i=1/n$ we get

$$ \frac{1}{n}\sum \log x_i \le \log \left( \frac{\sum x_i}{n}\right) $$ or

$$\log\left (\frac{x_1+ \ldots + x_n}{n} \right) \geq \frac{ \log x_1 +\log x_2 +\cdots \log x_n}{n}$$

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Draw graph of $y= \ln x$.

Choose $n$ points on it.A polygon of $n$ sides is thus formed whose coordinates of whose centroid is $$G \left (\frac{(x_1+x_2+....x_n)}n, \frac{(y_1 + .....y_n)}n\right ) = \left (\frac{(x_1+x_2+....x_n)}n, \frac{(\ln x_1 + \ln x_2 + .... + \ln x_n)}n\right )$$ Form centroid we draw a perpendicular to $x$ axis and extend it to meet the graph of $y= \ln x$ at $C$ and $y$ axis at $D$.

Now coordinates of point $C$ are $\left (\frac{(x_1+x_2+....x_n)}n, \ln \left (\frac{x_1+x_2+....x_n}n\right )\right )$. Now $$CD= \ln {\left (\frac{x_1+x_2+....x_n}n\right )}$$ ($ y$ coordinate of $C$). $$GD = \frac{\ln x_1 + \ln x_2 + .... + \ln x_n}n = \frac{\ln (x_1\cdot x_2\cdot ....x_n)} n$$ ($y$ coordinate of $G$). Now $CD> GD$ (as centroid of a convex polygon is always inside the polygon) $$\therefore \ln {\left (\frac{x_1+x_2+....x_n}n\right )} > \frac{\ln (x_1\cdot x_2\cdot ....x_n)} n = \ln \left ((x_1 \cdot x_2 \cdot ... x_n)^{\frac 1n} \right )$$ Now base and the number are positive. So we take antilog of both sides and get$$\frac{x_1+x2+....x_n}n > (x_1\cdot x_2\cdot ....x_n)^{1/n}$$ Now, it is also possible that all the selected $n$ points are only one point. Hence then the centroid is the point itself. Hence $\frac{x_1+x_2+....x_n}n = (x_1\cdot x_2\cdot ....x_n)^{1/n}$. Thereby proving that $\frac{x_1+x_2+....x_n}n \geq (x_1\cdot x_2\cdot ....x_n)^{1/n}$, where equality holds when$ x_1=x_2=...= x_n$.

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Can you please fix the formatting. – John Ma Jun 7 '15 at 3:33

Let $S(n)$ denote the statement $$ S(n):\; \frac{x_1+x_2+\cdots+x_n}{n}\geq\sqrt[n]{x_1x_2\ldots x_n},\quad n\in\mathbb{N}. $$ Base step ($n=1$): The statement $S(1)$ says that $\frac{x_1}{1}\geq\sqrt[1]{x_1}$, which is true because $x_1 = x_1$.

Base step ($n=2$): The statement $S(2)$ says that $$ \frac{x_1+x_2}{2}\geq\sqrt{x_1x_2},\tag{1} $$ which is true because $$ a\leq x \leq b \longleftrightarrow a+b\geq x+\frac{ab}{x}, \qquad 0<a\leq b,\; x>0 $$

Inductive step ($S(k)\to S(k+1)$): Fix some $k\geq 1$, where $k\in\mathbb{N}$. Assume that $$ S(k):\; \frac{x_1+x_2+\cdots+x_k}{k} \geq \sqrt[k]{x_1x_2\ldots x_k} $$ holds. To be proved is that $$ \frac{x_1+x_2+\cdots+x_{k+1}}{k+1}\geq\sqrt[k+1]{x_1x_2\ldots x_{k+1}} $$ follows. If $x_1 = x_2 = \cdots = x_{k+1}$, then the proof is done. If not, let $x_1x_2\ldots x_{k+1} = \rho^{k+1}$. Without loss of generality, assume that $x_1\leq x_i$ and $x_i \leq x_2$ for all $i$; that is, assume that $x_1 < \rho < x_2$. Beginning with the left side of $S(k+1)$ [excluding the $k+1$ divisor], \begin{align} x_1+x_2+\cdots+x_{k+1} &> \rho+\frac{x_1x_2}{\rho}+x_3+\cdots+x_k+x_{k+1}\tag{by $(1)$}\\ &\geq \rho+k\cdot\left(\sqrt[k]{\frac{x_1x_2}{\rho}x_3\ldots x_{k+1}}\right)\tag{by $S(k)$}\\ &= (k+1)\rho, \end{align} one arrives at the right side of $S(k+1)$ [with a $k+1$ multiple], thereby showing that $S(k+1)$ is also true, completing the inductive step. Thus, by mathematical induction, $S(n)$ is true for all $n\geq 1$, where $n\in\mathbb{N}$.

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