Proofs of AMGM inequality
Solution 1
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.
Consider for simplicity the twovariable case $(a+b)/2 \ge \sqrt{ab}$ and fix, say $a = 1$. The plot of $(1+b)/2$ and $\sqrt{b}$ show intuitively how the concave nature of the geometric mean implies it will always lie below the arithmetic mean. Also observe the equality at one point. In fact, this concavity will extend to any number of variables, but obviously a plot is not a proof.
The proof for more than two variables presented requires elementary properties of logarithms which transforms dealing with multiplication to dealing with addition. The inequality is rewritten in terms of logarithms:
$$ \frac{x_1 + \dots + x_n}{n}\ge (x_1 \dots x_n)^{1/n} $$
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), the inequality holds.
Original post of Pólya's Proof, using similar ideas of convexity of $e^x$:
Let $f(x) = e^{x1}x$. The first derivative is $f'(x)=e^{x1}1$ and the second derivative is $f''(x) = e^{x1}$.
$f$ is convex everywhere because $f''(x) > 0$, and has a minimum at $x=1$. Therefore $x \le e^{x1}$ 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
$$\sqrt[n]{x_1x_2\cdots x_n} \le a$$
Solution 2
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 twovariable 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?
Solution 3
LEMMA. If $a_1,\dots,a_n$ are positive numbers whose product is equal to $1,$ then $a_1+\dots+a_n\ge n,$ with equality only when $a_1=\cdots=a_n=1.$
Proof by induction on $n.$ The case $n=1$ being trivial, we suppose $n\ge2.$ Let $a_1,\ a_2,\ \dots,\ a_n$ be positive numbers with $a_1a_2\dots a_n=1.$ Without loss of generality, we can assume that $a_1=\max\{a_1,\dots,a_n\}\ge1$ and $a_2=\min\{a_1,\dots,a_n\}\le1.$ Thus we have $$a_1+a_2a_1a_21=(a_11)(1a_2)\ge0,$$ i.e., $$a_1+a_2a_1a_2\ge1.\tag1$$ Since $a_1a_2,\ a_3,\ a_4,\ \dots,\ a_n$ are $n1$ positive numbers with product equal to $1,$ by the inductive hypothesis we have $$a_1a_2+a_3+\cdots+a_n\ge n1.\tag2$$
Adding the inequalities (1) and (2), we get
$$a_1+a_2+a_3+\cdots+a_n\ge1+(n1)=n.\tag3$$
If equality holds in (3) then we must have equality in (1), that is, $a_1=1$ or $a_2=1,$ whence $a_1=\dots=a_n=1.$
THEOREM. If $x_1,\ x_2,\ \dots,\ x_n$ are positive numbers, then
$$\frac{x_1+x_2+\cdots+x_n}n\ge(x_1x_2\cdots x_n)^{\frac1n},$$
with equality only when $x_1=x_2=\cdots=x_n.$
Proof. Let $g=(x_1x_2\cdots x_n)^{\frac1n}.$ According to the lemma we have $$\frac{x_1}g+\frac{x_2}g+\cdots+\frac{x_n}g\ge n,$$ i.e., $$\frac{x_1+x_2+\cdots+x_n}n\ge g=(x_1x_2\cdots x_n)^{\frac1n}.$$ Equality holds only if $\frac{x_1}g=\frac{x_2}g=\dots=\frac{x_n}g=1,$ that is, $x_1=x_2=\cdots=x_n.$
Solution 4
As requested by dani_s, I will give the thermodynamic proof of the AMGM 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 $$\Delta S=c \ln{\frac{T}{T_0}}$$ where $c$ 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 proves the AMGM inequality.
Solution 5
As $(\sqrt{x_1}\sqrt{x_2})^2 \geq 0$ we have $$\sqrt{x_1 \cdot x_2} \leq \frac{x_1+x_2}{2}.$$ 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
$$\bigg( x_1 \ldots x_n A^{2^kn} \bigg)^{\frac{1}{2^k}} \leq \frac{x_1+\ldots+x_n+(2^kn) A}{2^k} = A.$$
Hence,
$$(x_1 \ldots x_n)^{1/n} \leq A = \frac{x_1+\ldots+x_n}{n}.$$
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Michiel Van Couwenberghe
I am a PhD student in mathematics at Ghent University (Belgium). My main research interest is non associative algebras and their connections to finite group theory, linear algebraic groups and incidence geometry.
Updated on October 30, 2021Comments

Michiel Van Couwenberghe over 1 year
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 nontrivial 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.

chubakueno about 9 yearsWhat have you seen, so that we don't repeat what you already know?

Hayden about 9 yearsAoPS has some interesting ones: artofproblemsolving.com/Wiki/index.php/Proofs_of_AMGM

Michiel Van Couwenberghe about 9 yearsInduction, backward induction, use of Jensen inequality, swapping terms, use of Lagrange multiplier, proof using thermodynamics are some of the proofs I know. Unfortunately, the proofs suggested by Hayden are also already on my list.

chubakueno about 9 yearsWould you include that in your post? (And,thermodynamics? ._. )

dani_s about 9 yearsI'm curious about the thermodynamics proof $\ddot \smile$

abnry about 9 yearsCommunity wiki?

vonbrand about 9 yearsPlease add the proofs you know as answers (just to get them all together).

Jose Antonio about 9 years

ASB about 8 yearspossible duplicate of Proving the AM:GM inequality

Admin about 8 years@aruna This version is more comprehensive, better to close the other one as a duplicate of this. Age of questions doesn't matter.


MartínBlas Pérez Pinilla about 9 yearsAlmost all your inequalities are reversed.

qwr about 9 years@MartínBlasPérezPinilla Woops, fixed

Siméon almost 9 yearsI believe this one is due to Cauchy.

Ragib Zaman almost 9 yearsThis 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.

Admin almost 8 yearsCan you please fix the formatting.

Mariano SuárezÁlvarez almost 8 yearsThere is no need to argue whether this is or not a rigorous proof, as it isn't.

DanielWainfleet over 7 yearsThe assumption is that entropy is minimum when all $T_i$ are equal, is known to be true only because of AGM is known to hold.

Saikat over 7 yearsI enjoyed this proof a lot. There were so many nice elements. I like how the product is always positive, and how nicely the a_1a_2 terms cancel each other out when you add the inequalities, and how elegantly this lemma is applied to prove the problem. Did you come up with this on your own?

ghosts_in_the_code over 6 yearsHow did you get step (2)? Shouldn't it be $a_1+a_2+a_3+\cdots$?

ghosts_in_the_code over 6 yearsOh ok, u've considered $a_1a_2$ as a single number, not as two distinct ones. Got it.

hukachaka over 5 years@qwr : i know this thread is a few years old but i still have a question. How is exp(xi/a  n)=exp(nn) ?

qwr over 5 years@hukachaka $a$ is defined as the arithmetic mean.

Jack D'Aurizio over 5 yearsPlease avoid posting duplicate answers: math.stackexchange.com/a/2527699/44121 You have already been contacted by a moderator in such regard.

Charlie Tian about 5 yearsinteresting and fun concept, but nonetheless very circular

Charlie Tian about 5 yearsmy favorite proof. utilizing the fact that this is a corollary of the behavior of convex functions and tying it together is much better IMO than ad hoc algebraic manipulations

tortue almost 5 yearsI think Hölder's inequality is way more general than AMGM and proving the latter through the former is something artificial.

Bob almost 5 years@pointguard0 If that is your problem, you can always use Cauchy Schwarz to get the result for $p=1/2,q=1$, then $p=1/4, q=1/2$, and so on...

Sandeep Silwal over 3 yearsVery nice proof! :)

CopyPasteIt about 3 yearsI wanted an elementary algebraic proof (with the examination of rectangles/squares in 2dim driving it) which leads to $\quad \sum_{i=1}^n x_i^n \ge n \prod_{i=1}^n x_i$. $\quad$ But this would inexorably lead to proving your lemma! Although dependent on 'proof aesthetics', can't understand why this is #1 answer. (+1)

CopyPasteIt about 3 yearsI wanted to extend the 2dim proof to higher dimension and prove $\quad \sum_{i=1}^n x_i^n \ge n \prod_{i=1}^n x_i$. $\quad$ The techniques employed in bof's answer gets the job done.

Batominovski almost 3 yearsPlease copy your answer here to this link: math.stackexchange.com/questions/3798002/…. In a week, I may copy your answer into the link so that the answer is preserved. The answer will be a Community Wiki post, and you can remove the part that belongs to you, and put it in your own separate answer. (I'm not sure if you were notified from my comment in the deleted post.)

Philipp about 1 yearThe fact that $a_1=a_2=1$ is clear but I don't see why you can conclude $a_1=\dots=a_n=1$? It would be possible to have, say $a_3<1$ and $a_4>1$ and still satisfy $a_1+a_2+a_3+\cdots+a_n=n$. Can you elaborate a bit?

bof about 1 year@Philipp I assumed that $a_1=\max\{a_1,\dots,a_n\}$ and $a_2=\min\{a_1,\dots,a_n\}$.