# How to understand Cauchy's proof of AM-GM inequality(the last step)

The AM-GM inequality: $$a_1a_2\cdots a_n\leq\left(\frac{a_1+\cdots + a_n}{n}\right)^n$$

the trivial case: $a_1a_2 \leq \left(\frac{a_1+a_2}{2}\right)^2$ is self-evident.

then cauchy use this fact repeatedly. He get: $$a_1a_2\cdots a_{2^m} \leq \left(\frac{a_1+a_2+\cdots +a_{2^m}}{2^m}\right)^{2^m}.\tag{\ast}$$

This is easy to understand, and natural, but next step:

he let $$b_1 = a_1,\ b_2=a_2,\ \ldots,\ b_n=a_n,\ b_{n+1}=\cdots=b_{2^m}=\frac{a_1+a_2+\cdots a_n}{n}$$

replace $a_1, a_2,\ldots ,a_{2^m}$ with $b_1, b_2, \ldots b_{2^m}$ in $(\ast)$ and simplified both left and right. He get the answer. I can repeat his proof, but I don't know why Cauchy can got this idea. How does he got it?

• I used to know this...
– leo
Apr 4 '14 at 17:31

You proved that $$a_1a_2\cdots a_{2^m} \leq \left(\frac{a_1+a_2+\cdots +a_{2^m}}{2^m}\right)^{2^m}\qquad\forall m\in \Bbb N, \tag{\star\star}$$ and you want to prove $$a_1a_2\cdots a_n\leq\left(\frac{a_1+\cdots + a_n}{n}\right)^n\qquad\forall n\in\Bbb N\tag{\star}.$$

You have stablished $$(\star)$$ for $$n=2$$ so you have to prove it now for all $$n\gt 2$$. So pick $$n\gt 2$$ and $$n$$ numbers $$a_1,\ldots,a_n$$.

Now, you can use $$(\star\star)$$ for a list of numbers of size a power of two. You don't know if $$n$$ is a power of 2 but you know that sure there is a $$m$$ so that $$2^m\gt n$$, so let's add some numbers to our list so we end with a list of size $$2^m$$.

Which numbers should we add? We don't know, but the most simple thing we can do is add the same number several times (after all we're trying, we don't know if it will work), then our list will be $$a_1,\ldots,a_n,\underbrace{x,\ldots,x}_{2^m-n}$$.

Using $$(\star\star)$$ we get $$a_1\cdots a_n\cdot x^{2^m-n}\leq \left(\frac{a_1+\cdots+(2^m-n)x}{2^m}\right)^{2^m}.\tag{\ast}$$

Now you wonder, it would be nice if $$x$$ had the ability of turn$$\frac{a_1+\cdots+a_n+(2^m-n)x}{2^m}$$ into $$\frac{a_1+\cdots+a_n}{n}$$,is there such $$x$$? That question is answered by the equation $$\frac{a_1+\cdots+a_n+(2^m-n)x}{2^m}=\frac{a_1+\cdots+a_n}{n},$$ which by solving for $$x$$ tells you that $$x=\frac{a_1+\cdots+a_n}{n}$$ work.

And it turns out that this $$x$$ does the magic to turn $$(\ast)$$ into $$(\star\star)$$. I don't know if Cauchy got it this way, but it's a possible path.

I don't know what Cauchy was thinking, but let me throw some ideas at each other.

When you're trying to prove AM/GM, one natural idea is to think of operations on sequences that preserve their arithmetic mean and to see what they do to the geometric mean (or vice versa). (See Dijkstra's proof, for example.) If you're also thinking you might prove it by induction, you might especially consider operations that add one term to the sequence, or take one away. Putting those two ideas together, you notice that the arithmetic means of $a_1,\dotsc,a_n$ and of $a_1,\dotsc,a_n,\frac{a_1+\dots+a_n}{n}$ are obviously equal. Then you write down the AM/GM inequality for the extended sequence, do some simplifications, and you find that you've proved $$\text{AM/GM for a_1,\dotsc,a_n,\tfrac{a_1+\dots+a_n}{n}} \iff \text{AM/GM for a_1,\dotsc,a_n}$$ and of course, by induction, $$\text{AM/GM for a_1,\dotsc,a_n,\underbrace{\tfrac{a_1+\dots+a_n}{n},\dotsc,\tfrac{a_1+\dots+a_n}{n}}_{m \text{ terms}}} \iff \text{AM/GM for a_1,\dotsc,a_n}$$ At this point you become slightly annoyed because this is exactly what you'd need if your induction went backwards. But then you realize...

His idea is to prove it by mathematical induction, albeit in a different form. Let $\{P(n): n \in \mathbb{N}\}$ be a sequence of propositions.

Key idea: If $P(n)$ is true for infinitely many $n$ and $P(n) \implies P(n-1)$ for all $n$, then $P(n)$ is true for all $n$.

Let $P(n)$ be the A.M-G.M. inequality for $n$ numbers. Cauchy proves it for infinitely many $n$, namely the powers of 2. He then proves the implication $P(n) \implies P(n-1)$.

I don't know how Cauchy found it - perhaps he first noticed that it is easy to prove $P(n) \implies P(n-1)$ than to prove $P(n-1) \implies P(n)$.

• That proof is different from the one OP is talking about.
– leo
Apr 4 '14 at 17:52
• The proof you mention use this kind of induction.
– leo
Apr 4 '14 at 19:02