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***It’s recommended that you read this article on a computer or tablet. I apologize that the formatting isn’t great on a smartphone!***

One of my favorite theorems (inequalities) is the AM-GM inequality (or simply the AM-GM). “Why?” I hear you ask through the screen, it’s because there are so many wonderful proofs of the darn thing! From a beautiful geometric proof, to a creative `two steps forward, one step back‘ induction due to Cauchy, to a more analytic proof due to Pólya. I’m always pleasantly surprised by how many ways there are to prove the AM-GM. Today, we will cover these aforementioned proofs, so I hope that you have as much fun as I did when I learnt them!

Oh, and just so you know, and don’t get confused, I use the symbol below to denote the end of a proof.

Introduction

Each of the proofs we give will generalize our AM-GM inequality. This is one of the reasons why I find the AM-GM inequality to be so amazing. As we will see, each proof is creative and enlightening. However, each proof is also restrictive. For example, the geometric proof of the special case of the AM-GM won’t be very useful for proving the standard AM-GM inequality. And even though the proof we give for the standard AM-GM, due to Cauchy, can be generalized to get at the general AM-GM, the process is not that fun (in my opinion). Especially when we compare it to Pólya’s proof. So, as we climb the AM-GM ladder of generality, I hope you find some of the ideas enjoyable!

P.S. I apologize to my Master’s Thesis advisor (and Real Analysis 2 professor) Professor Joris Roos for breaking one of his rules, which is: do not use dots ($\cdots$) in proofs when trying to capture patterns. I will be doing a lot of this when writing $a_1+\cdots + a_n$ instead of $\sum_{k=1}^na_k$. Again, very sorry.

The Special Case AM-GM (by Geometry)

We begin with the following inequality:

Theorem (Special Case AM-GM): Let $a,b\in \R$ be non-negative. Then,

$$\sqrt{ab}\leq \frac{a + b}{2}.$$

A wonderful inequality that follows quickly from the algebraic fact that $ab = \frac{1}{2}\Big(a^2 + b^2 -(a-b)^2\Big).$ However, there is a more geometric proof that is beautiful. This is our first proof.

Proof of Special Case AM-GM : (Click to see proof)

Our goal is to determine what $h$ is equal to in the following figure:

I challenge you to try to figure it out! As a hint, you will need to prove something about some of the angles in the triangles…

Something about angles: (Click to see Lemma)

As it happens, we want to show that $\angle ch$ is a right angle.

To do so, consider the following figure:

You might notice that we have two isosceles triangles (each has a radius as two of its legs). From this, we deduce: $2\beta+2\theta = 180\degree,$ which implies $\beta+\theta = 90\degree.$

Remark: We can also deduce the following as well: $\alpha+ \phi = 180\degree,\;\;2\theta + \phi = 180\degree.$ From these equations, we can see that $\alpha =2\theta,$ which you may recall comes from the inscribed angle theorem.

In summary we have:

Hi!

Proof: (Click to see Full Proof)

Perfect, we have the following:

The key step is to notice that $\triangle ahc$ and $\triangle hbd$ are similar triangles (which follows from the fact that they both have the same internal angles, can you see why?). Thus, $a/h = h/b$ from which we deduce $h = \sqrt{ab}.$

The final steps are to note that $h \leq radius$, for any triangle that we draw, and that $radius = \frac{a+b}{2}.$

Hi!

Hi!

How nice is that proof?

Cauchy’s Induction and the Standard AM-GM

We now take our first step forward and prove what is usually called the AM-GM inequality:

Theorem (Standard AM-GM Inequality): Let $a_1,a_2,\cdots,a_n\in \R_{\geq0}.$ Then,

$$\sqrt[n]{a_1a_2\cdots a_n}\leq \frac{a_1 + \cdots +a_n}{n}.$$

This proof by induction is, in my opinion, striking the first time you see it. I hope you enjoy the two steps forward, one step back approach of this proof.

Scratch Work:

We know that we will use induction on $n.$ The base case $n=1$ is trivial, and we already showed the first nontrivial case $n=2$ above. After some experimentation, we might get stuck on the inductive step because it is not easy to show that the AM-GM for $n$ implies the AM-GM for $n+1.$ But, if we are lucky, we might notice that we can apply the $n=2$ case twice to get the $n=4$ case,

$(a_1a_2a_3a_4)^{1/4} = ((a_1a_2)(a_3a_4))^{1/4} \\ \qquad\qquad\qquad\leq \frac{(a_1a_2)^{1/2}+(a_3a_4)^{1/2}}{2}\\ \qquad\qquad\qquad \leq \frac{\frac{a_1+a_2}{2}+\frac{a_3+a_4}{2}}{2}\\ \qquad\qquad\qquad = \frac{a_1+a_2+a_3+a_4}{4}.$

As we will see, we can use this same idea in our induction to prove $n=2^{k+1}$ using $n=2^k.$ However, this strategy only proves the AM-GM for $n$ equal to a power of two. How do you think we can get every $n$ from this?

The Proof

Proof of AM-GM: (Click to see proof)

We will first use induction to prove the AM-GM for powers of two. That is, we will induct on $k$ where $n=2^k.$ Our base case $n=2^1 = 2$ has already been done above. Now assume that the AM-GM inequality holds for $n=2^k$ and consider $n=2^{k+1}.$ Observe the following,

$(a_1a_2\cdots a_{2^k} a_{2^k +1}\cdots a_{2^{k+1}})^{1/2^{k+1}} \leq \frac{(a_1a_2\cdots a_{2^k})^{1/2^k} + (a_{2^{k}+1}a_{2^{k}+2}\cdots a_{2^{k+1}})^{1/2^k} }{2} \\\qquad\qquad \qquad \qquad\qquad\qquad\qquad\;\;\;\leq \frac{a_1 + \cdots +a_{2^{k+1}}}{2^{k+1}}.$

Where we used both our $n= 2$ base case and our induction hypothesis on $(a_1a_2\cdots a_{2^k})^{1/2^k}$ and $(a_{2^k+1}a_{2^k+2}\cdots a_{2^{k+1}})^{1/2^k}.$ This closes the induction on our powers of two.

Before we move on, what did you come up with? If you got it, leave a comment!

We will show that the AM-GM inequality holds for all $n.$ We do this by demonstrating that the $2^k>n$ case implies the $n$ case. How might we do this? Take a moment and see what you come up with!

The idea is to define a new sequence $b_1,b_2,\cdots,b_{2^k}$ that is equal to

$$b_i = \begin{cases} a_i,\;\;\;\;\;\;\;\;\mathrm{if}\;1\leq i\leq n\\\;\; \;A,\;\;\;\;\mathrm{if}\;n+1\leq i\leq 2^k. \end{cases}$$

Where $A:= \frac{a_1+a_2+\cdots+a_n}{n}.$ Now we can use the AM-GM for $n=2^k$ to deduce,

$$\begin{align} (b_1b_2\cdots b_{n} b_{n +1}\cdots b_{2^{k}})^{1/2^{k}} \leq \frac{b_1 + \cdots +b_{2^{k}}}{2^{k}} \\ \textit{ }\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;=\frac{a_1 + \cdots +a_{n} + (2^k-n)A}{2^{k}}\\ \textit{ }\qquad\qquad\qquad\qquad\;\;\;=\frac{2^kA}{2^{k}} \end{align}$$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;=A .\qquad\qquad\qquad\qquad\qquad\qquad(\char”2705)$

Simplifying the $(b_1b_2\cdots b_{n} b_{n +1}\cdots b_{2^{k}})^{1/2^{k}}$ we get

$$\begin{align} (b_1b_2\cdots b_{n} b_{n +1}\cdots b_{2^{k}})^{1/2^{k}} =(a_1a_2\cdots a_{n} A^{2^{k}-n})^{1/2^{k}} \\\qquad\qquad\qquad\;\qquad\qquad\qquad\;\qquad\qquad\qquad\;\;\;\;\;= \frac{(a_1a_2\cdots a_{n})^{1/2^{k}}A}{A^{n/2^k}} .\qquad\qquad\qquad\qquad(\char”26A1) \end{align}$$

It follows from $(\char”2705)$ and $(\char”26A1)$,

$$\begin{align} \frac{(a_1a_2\cdots a_{n})^{1/2^{k}}}{A^{n/2^k}} \leq 1. \end{align}$$

Moving $A^{n/2^k}$ to the right we get

$$(a_1a_2\cdots a_{n})^{1/2^{k}} \leq A^{n/2^k}.$$

Which we then raise to the $2^k/n$ th power to get our final result:

$$(a_1a_2\cdots a_{n})^{1/n} \leq A = \frac{a_1+a_2+\cdots+a_n}{n}.$$

Concluding the proof.

A modest Generalization

Before moving on to George Pólya’s proof of the most general AM-GM inequality that we will discuss, we can actually generalize our result to the following,

Theorem (Modestly General AM-GM): Let $a_1,a_2,\cdots,a_n \in \R$ be nonnegative and $p_1,p_2,\cdots,p_n\in \mathbb{Q}$ be positive rational numbers such that $p_1+p_2+\cdots+p_n=1.$ Then,

$$a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}\leq p_1a_1 + p_2a_2\cdots +p_na_n.$$

Remark: Note that when $p_1=p_2 = \cdots = p_n = 1/n,$ we recover the standard AM-GM inequality.

Our most general AM-GM will allow for $p_1,p_2,\cdots,p_n$ to be any positive sequence of real numbers that satisfy $p_1+p_2+\cdots+p_n=1.$ However, we can already tackle the case where $p_1,p_2,\cdots,p_n\in \mathbb{Q} .$ To do so, we will use a similar trick that we used in Cauchy’s proof. Namely, we will define a new sequence that contains multiple copies of elements from $a_1,a_2,\cdots,a_n.$

Suppose that $M\in \N$ is large enough so that, for all $i$, we have $p_i = \frac{q_i}{M}$ for some $q_i\in \N.$ Now consider the product

$\Bigg(\underbrace{(a_1^{1/M} a_1^{1/M}\cdots a_1^{1/M})}_{q_1-\mathrm{times}} \cdots \underbrace{(a_n^{1/M} a_n^{1/M}\cdots a_n^{1/M})}_{q_n-\mathrm{times}} \Bigg)\,^{1/M} \leq \frac{1}{M}\Bigg( q_1a_1 + q_2a_2\cdots +q_na_n \Bigg) \\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;=p_1a_1 + p_2a_2\cdots +p_na_n.$

BOOM!

George Pólya’s Proof

The proof we are going to give is one that the great mathematician George Pólya discovered. He said that this proof came to him in a dream, which is very impressive! This proof is like a good movie; there is suspense and a twist at the end. I hope you enjoy it!

Here we go!

This proof relies on the following observation/proposition:

Proposition: For any $x\in \R,$ we have $1+x \leq e^x.$

However, we will require a quick lemma before “we” prove the above proposition.

Note: If you are okay with taking this proposition as a fact based on the graph above, then you can jump to the proof of the General AM-GM.

Lemma (Bernoulli’s Inequality): For all $x\in [-1,\infty)$ and $n\in \N$ we have

$1+nx \leq (1+x)^n.$

Proof of Lemma: (Click to see proof)

Let $x\in [-1,\infty).$ We will prove Bernoulli’s inequality by induction on $n.$ Our base case is immediate: $1+x \leq (1+x)^1.$ Now assume that $1+nx \leq (1+x)^n$ for some $n\in \N$ and consider $1+(n+1)x.$ This induction step is one of those cases where we discover what we need to do by working backwards until we get what we want, and then when we type out the proof, we write it forwards, obscuring how we might have come up with the argument. However, I hope that you will allow me to simply give the tidy proof without the justification (in fact, it might be a fun challenge to see if you can figure out how you might come up with it!)

Observe the following steps:

$1+(n+1)x \leq 1+(n+1)x +nx^2\qquad \mathrm{because}\; nx^2\geq0 \\ \qquad\qquad\;\;\;\;\;= 1+nx+x +nx^2 \\ \qquad\qquad\;\;\;\;\;=(1+nx)(1+x)\\ \qquad\qquad\;\;\;\;\;\leq(1+x)^n(1+x)\\ \qquad\qquad\;\;\;\;\;=(1+x)^{n+1}.$

Closing the induction and completing the proof.

Concluding the proof.

*Where did we use the fact that $x\in [-1,\infty)?$ We used it when we said: $(1+x)^n (1+x) \geq (1+nx)(1+x).$ This is because $x\in [-1,\infty)$ implies $1+x\geq 0.$

Proof Idea of Proposition: (Click to see proof)

Let $x\in \R.$ First, note that for $x\lt-1$, we have $1+x \lt0 \lt e^x.$ Next, for $x\geq -1$ we can use Bernoulli’s inequality. Recalling that $e^x$ (Euler’s number) is given by the following limit:

$$e^x= \lim_{n\rightarrow \infty}{(1+\frac{x}{n})^n},$$

and noting that $\Big((1+\frac{x}{n})^n\Big)\,_{n\in \N}$ is a monotone increasing sequence (or simply that $(1+x/n)^n \leq e^x$ for all $n$ and $x$) while letting $x/n \mapsto x$ in Bernoulli’s inequality, gives the desired result. So that you have something fun to do, I will leave the details of this proof to you! You’re welcome!

Concluding the proof.

Proof of the AM-GM

Now on to the fun part!

Theorem (Generalized AM-GM): Let $a_1,a_2,\cdots,a_n\in \R$ and $p_1,p_2,\cdots,p_n \in \R$ be nonnegative. If the sequence $(p_j)$ are such that $p_1+p_2+\cdots+p_n=1.$ Then,

$a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}\leq p_1a_1 + p_2a_2\cdots +p_na_n.$

Where we get equality if and only if $a_i= a_j$ for all $i,j.$

Proof (Beginning): Let $a_1,a_2,\cdots,a_n\in \R$ and $p_1,p_2,\cdots,p_n \in \R$ be nonnegative such that $p_1+p_2+\cdots+p_n=1.$

First, consider the situation where $a_1,a_2,\cdots,a_n=0.$ Indeed, $0\leq0$ (yup, that is the easiest inequality you will ever see), so the AM-GM is trivially satisfied.

With that out of the way, notice that $x\leq e^{x-1}$ by our proposition. Furthermore, $x^k\leq (e^{x-1})^k = e^{kx-k}.$ Hence,

$a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}\leq (e^{p_1a_1-p_1})\;(e^{p_2a_2-p_2})\;\cdots (e^{p_na_n-p_n})\\\qquad \qquad \;\;\;\;\;\,=e^{p_1a_1+p_2a_2 + \cdots p_na_n-(p_1+p_2+ \cdots p_n)}\\ \qquad \qquad \;\;\;\;\;\,=e^{p_1a_1+p_2a_2 + \cdots p_na_n-1} \qquad \mathrm{since}\;p_1+p_2+\cdots+p_n=1.\\\qquad \qquad \;\;\;\;\;\,=e^{\Big(\sum_{i=1}^np_ia_i\Big)-1}$

We now have an upper bound for $a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}.$ How might we go on from here? This is where our twist comes in! We will normalize our sequence $a_1,\cdots,a_n$ to obtain the desired result. See what you might come up with before you continue reading!

Proof: (Click to see proof)

First, note that the inequality $a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}\leq e^{\Big(\sum_{i=1}^np_ia_i\Big)-1}$ holds for any nonnegative sequence. This means we can use the result on a new sequence!

Let’s consider the sequence $b_1,b_2,\cdots,b_n \in \R$ where $b_1 =a_1 /A\geq 0$ and $A =\sum_{i=1}^np_ia_i= p_1a_1 + p_2a_2\cdots +p_na_n.$ Using what we’ve already established,

$\Big(\frac{a_1}{A}\Big)\,^{p_1}\cdot \Big(\frac{a_2}{A}\Big)\;^{p_2}\cdots \,\Big(\frac{a_n}{A}\Big)\,^{p_n}={b}_1^{p_1}b_2^{p_2}\cdots b_n^{p_n} \\\qquad\qquad\qquad\qquad\qquad\qquad\;\leq e^{\Big(\sum_{i=1}^np_ib_i\Big)-1}\\\qquad\qquad\qquad\qquad\qquad\qquad\;=e^{\Big(\sum_{i=1}^np_ia_i\Big)/A-1} \\\qquad\qquad\qquad\qquad\qquad\qquad\;= e^{1-1} \\\qquad\qquad\qquad\qquad\qquad\qquad\;= e^0 \\\qquad\qquad\qquad\qquad\qquad\qquad\;= 1.$

It follows that,

$$\frac{a_1^{p_1}\cdot a_2^{p_2}\cdots a_n^{p_n}}{A^{p_1+\cdots + p_n}}\leq 1.$$

Recalling $p_1+p_2+\cdots+p_n=1,$we deduce

$$a_1^{p_1}\cdot a_2^{p_2}\cdots a_n^{p_n}\leq A = p_1a_1 + \cdots +p_na_n,$$

almost concluding the proof. The last thing we need to do is argue why we have equality if and only if $a_i= a_j$ for all $i,j.$

Note that $\frac{a_i}{A}\lt e^{a_i/A-1}$ unless $\frac{a_i}{A} =1.$ Thus, we have a strict inequality unless $a_i=A$ for all $i.$ Now we are done the proof.

Concluding the proof.

How Wonderful!

I find all three proofs to be absolutely wonderful. Each requires some creativity, and each is completely different. In Cauchy’s proof, we can see how creative we can be when using induction. And in Pólya’s proof, we need to notice the key inequality $1+x \leq e^x,$ as well as normalization. Pólya’s proof is the first place where I saw the idea of normalization used in such a crucial way. Normalization is a great tool to have in your mathematical toolbox, so I hope that Pólya’s proof thrilled you (and taught you about) the concept of normalization as much as it did for me when I first learnt it.

A gift for you

I hope that you had some fun with these today, but I can’t leave you empty-handed now, can I? So, in an effort to give you something to work on, I challenge you to prove the AM-GM-HM inequality:

Theorem (Standard AM-GM-HM Inequality): Let $a_1,a_2,\cdots,a_n\in \R_{\geq0}.$ Then,

$$\frac{a_1 + \cdots +a_n}{n} \geq \sqrt[n]{a_1a_2\cdots a_n}\geq \frac{n}{\frac{1}{a_1} + \cdots +\frac{1}{a_n}}.$$

Note that all you need to show is the right-most inequality with the harmonic mean (HM).

Click if you want a hint.

Use the standard AM-GM inequality with the sequence $(1/a_i)_i.$

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P.S. Thank you for reading (at least part of) this long article! Please leave a comment if you notice any typos or if you have suggestions for how I can improve my writing! Thank you in advance!