All content on this site is available under the Creative Commons Attribution 4.0 International license: Math Puzzle from March 1, 2019
Can you prove the famous AM-GM inequality?
Written on by Noble Mushtak

The following was a puzzle presented to Marshwood GT students on March 1, 2019. Have fun doing math!

It is well known that the $$\log$$ function is concave. In other words, for any real number $$0 \leq \lambda \leq 1$$ and any positive numbers $$x,y$$:

$$\log(\lambda x+(1-\lambda)y) \geq \lambda \log(x)+(1-\lambda)\log(y) \ (1)$$

For example, if we let $$\lambda=\frac{1}{2}$$, then we find that:

$$\log\left(\frac{x+y}{2}\right)\geq \frac{\log(x)+\log(y)}{2} \ (2)$$

On the other hand, if we let $$\lambda=\frac{1}{3}$$, we get:

$$\log\left(\frac{x+2y}{3}\right) \geq \frac{1}{3}\log(x)+\frac{2}{3}\log(y) \ (3)$$

As you can see, the fact that $$\log$$ is concave is very powerful, because it can help prove a lot of different inequalities. Now, use the fact that $$\log$$ is concave to prove the AM-GM inequality: For any group of $$n$$ positive numbers $$a_1, a_2, ..., a_n$$, the following holds true:

$$\frac{a_1+a_2+...+a_n}{n} \geq \sqrt[n]{a_1a_2...a_n} \ (4)$$

Part 0: Use logarithms to show that AM-GM inequality is equivalent to the following:

$$\log\left(\frac{a_1+a_2+...+a_n}{n}\right)\geq \frac{\log(a_1)+\log(a_2)+...+\log(a_n)}{n} \ (5)$$

From here on, we will focus on the proving inequality $$(5)$$ rather than the actual AM-GM inequality because they are equivalent, so proving inequality $$(5)$$ is the same as proving AM-GM inequality.

Inequality $$(5)$$ is also commonly known as the finite-form of Jensen's inequality.

Part 1: First, let's focus on a small value of $$n$$: If $$n=2$$, then the inequality becomes:

$$\log\left(\frac{a_1+a_2}{2}\right)\geq \frac{\log(a_1)+\log(a_2)}{2} \ (6)$$

Plug in $$\lambda=\frac{1}{2}$$ into inequality $$(1)$$ to prove inequality $$(6)$$.

Part 2: Now, let's make things a bit harder: If $$n=3$$, then the inequality becomes:

$$\log\left(\frac{a_1+a_2+a_3}{3}\right)\geq \frac{\log(a_1)+\log(a_2)+\log(a_3)}{3} \ (7)$$

Now, if we plug in $$x=a_1$$ and $$y=\frac{a_2+a_3}{2}$$ into inequality $$(3)$$, we get:

$$\log\left(\frac{a_1+a_2+a_3}{3}\right)\geq \frac{1}{3}\log(x)+\frac{2}{3}\log\left(\frac{a_2+a_3}{2}\right) \ (8)$$

Combine inequalities $$(6)$$ and $$(8)$$ in order to prove inequality $$(7)$$.

Part 3: Finally, let's try one more example: If $$n=4$$, then the inequality becomes:

$$\log\left(\frac{a_1+a_2+a_3+a_4}{4}\right)\geq \frac{\log(a_1)+\log(a_2)+\log(a_3)+\log(a_4)}{4} \ (9)$$

Now, the key thing to realize here is that:

$$\frac{a_1+a_2+a_3+a_4}{4}=\frac{1}{4}a_1+\frac{3}{4}\frac{a_2+a_3+a_4}{3}$$

Therefore, if we plug in $$\lambda=\frac{1}{4}, x=a_1, y=\frac{a_2+a_3+a_4}{3}$$ into inequality $$(1)$$, we get:

$$\log\left(\frac{a_1+a_2+a_3+a_4}{4}\right)\geq \frac{1}{4}\log(a_1)+\frac{3}{4}\log\left(\frac{a_2+a_3+a_4}{3}\right) \ (10)$$

Combine inequalities $$(7)$$ and $$(10)$$ in order to prove inequality $$(9)$$.

Part 4: Hopefully, you realize the pattern by now: In order to prove the inequality for $$n=k$$, you need to use the inequality for $$n=k-1$$. This is a really powerful technique called mathematical induction, and it allows us to prove AM-GM inequality for all positive integers $$n$$. First, we state that the inequality is true for $$n=k-1$$, assuming that it has already been proven:

$$\log\left(\frac{a_1+a_2+...+a_{k-1}}{k-1}\right)\geq \frac{\log(a_1)+\log(a_2)+...+\log(a_{k-1})}{k-1} \ (11)$$

Then, we use this inequality to prove the inequality for $$n=k$$. To be clear, the inequality for $$n=k$$ is the following:

$$\log\left(\frac{a_1+a_2+...+a_k}{k}\right)\geq \frac{\log(a_1)+\log(a_2)+...+\log(a_k)}{k} \ (12)$$

Now, just like in Part 3, a key thing to notice is that:

$$\frac{a_1+a_2+...+a_k}{k}=\frac{1}{k}a_1+\frac{k-1}{k}\frac{a_2+a_3+...+a_k}{k-1}$$

Thus, if we plug in $$\lambda=\frac{1}{k}, x=a_1, y=\frac{a_2+a_3+...+a_k}{k-1}$$ into inequality $$(1)$$, we get:

$$\log\left(\frac{a_1+a_2+...+a_k}{k}\right)\geq \frac{1}{k}\log(a_1)+\frac{k-1}{k}\log\left(\frac{a_2+a_3+...+a_k}{k-1}\right) \ (13)$$

Combine inequalities $$(11)$$ and $$(13)$$ to prove inequality $$(12)$$.

After proving inequality $$(12)$$, we can finish off our whole proof: By induction, since the AM-GM inequality for $$n=k-1$$ implies the AM-GM inequality for $$n=k$$, the AM-GM inequality must be true for all positive integers $$n$$. Q.E.D.

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