The Harmonic Mean and the AM-HM Inequality

Here's a proof of AM-HM using just AM-GM.

$$\frac{\frac1{a_1}+\dots+\frac1{a_n}}n\ge\sqrt[n]{\frac1{a_1}\dots\frac1{a_n}}$$ $$\frac{n}{\frac1{a_1}+\dots+\frac1{a_n}}\le\sqrt[n]{a_1\dots a_n}$$ $$\sqrt[n]{a_1\dots a_n}\le\frac{a_1+\dots+a_n}n$$

Inequalities (Part 3)

In this part, we'll look at more advanced applications of AM-GM.

First, we show the AM-GM-HM inequality, which is

$\frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1a_2...a_n} \ge\frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + ... + \frac{1}{a_n}}$, for all positive integers $n$ and positive real numbers $a_1, a_2, ..., a_n$. (The interested reader can try to prove this, it's not really hard.)

Consider this problem :

Prove that for all positive real $a, b, c, d$,

$\frac{a + c}{a + b} + \frac{b + d}{b + c} + \frac{a + c}{c + d} + \frac{b + d}{a + d} \ge 4$.

The idea is, we pair the first and third terms, as well as the second and fourth terms.

By AM-HM, we can write

$\frac{1}{a + b} + \frac{1}{c + d} \ge \frac{4}{a + b + c + d}$, $\frac{1}{b + c} + \frac{1}{a + d} \ge \frac{4}{a + b + c + d}$. Thus,

$\frac{a + c}{a + b} + \frac{b + d}{b + c} + \frac{a + c}{c + d} + \frac{b + d}{a + d} = (a + c)[\frac{1}{a + b} + \frac{1}{c + d}] + (b + d)[\frac{1}{b + c} + \frac{1}{a + d}] \ge (a + c)[\frac{4}{a + b + c + d}] + (b + d)[\frac{4}{a + b + c + d}] = 4 \blacksquare$

That was not too hard. We continue with $2$ more APMO problems.

(APMO 2009) Consider the following operation on positive real numbers written on a blackboard : choose a number $r$ on the blackboard, erase the number, an then write a pair of positive real numbers $a$ and $b$ satisfying $2r^2 = ab$ on the board. Assume that you start out with just one positive real number $r$ on the blackboard, and apply this operation $k^2 - 1$ times to end up with $k^2$ positive real numbers (not neccesarily distinct). Show that there exists a number on the board which does not exceed $kr$.

(APMO 1998) Let $a, b, c$ be positive real numbers. Prove that $(1 + \frac{a}{b})(1 + \frac{b}{c})(1 + \frac{c}{a}) \ge 2(1 + \frac{a + b + c}{\sqrt[3]{abc}})$.

The first problem was problem $1$ on APMO $2009$, but was exceptionally hard. It is clear that there's a monovariant here, but what?

Consider the sum of the squares of reciprocals of the numbers on board. In the beginning, the sum is $\frac{1}{r^2}$. At each move, after erasing $r$ and writing $a$ and $b$ changes the sum by

$\frac{1}{a^2} + \frac{1}{b^2} - \frac{1}{r^2} = \frac{a^2 + b^2}{a^2b^2} - \frac{1}{r^2} \ge \frac{2ab}{a^2b^2} - \frac{1}{r^2} = \frac{2}{ab} - \frac{1}{r^2} = 0.$ In other words, the sum is nondecreasing.

Suppose after $k^2 - 1$ moves, the numbers on the board are $x_1, x_2, ..., x_{k^2}$, with $x_1$ being the smallest. Then since the sum of squares of the reciprocals exceed $\frac{1}{r^2}$, we have

$k^2(\frac{1}{x_1^2}) \ge \frac{1}{x_1^2} + \frac{1}{x_2^2} + ... + \frac{1}{x_k^2} \ge \frac{1}{r^2}$, thus $x_1 \le kr$. $\blacksquare$.

The main idea (and I don't know how could anyone come up with it) is finding the correct monovariant, which is usually a tough thing to do. (as in the case with this problem)

Finally, we will look at the APMO 1998 problem.

First, we expand the LHS, to get $2 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{b}{a} + \frac{c}{b} + \frac{a}{c}$. Let the sum of the $6$ fractions be $S$. Now, we claim that $\frac{2}{3}S + 2 \ge 2(\frac{a + b + c}{\sqrt[3]{abc}})$ and $\frac{1}{3}S \ge 2$. Combining these $2$ gives the desired conclusion.

For the first claim, we prove that $\frac{S}{3} + 1 \ge \frac{a + b + c}{\sqrt[3]{abc}}$. Observe that

$\frac{1 + \frac{a}{b} + \frac{a}{c}}{3} \ge \frac{a}{\sqrt[3]{abc}}$ by AM-GM. Adding up similar inequalities gives the first claim.

For the second claim, we need to prove that $S \ge 6$, which follows from AM-GM. Thus, we're done. $\blacksquare$

Conclusion : AM-GM is very useful in Olympiad Maths. It almost always comes up.

The arithmetic mean and harmonic mean of a set of $n$ positive integers $x_1, x_2, \dots , x_n$ are respectively given by $$\frac{x_1+x_2 + \cdots +x_n}{n}$$ and $$\frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots \frac{1}{x_n}} .$$ The AM-HM inequality states that $$\frac{x_1+x_2 + \cdots +x_n}{n} \ge \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots \frac{1}{x_n}}.$$ Here is a proof: $$n^2 = \left(x_1+x_2 + \cdots + x_n \right) \left(\frac{n^2}{x_1+x_2 + \cdots + x_n} \right) \le \left(x_1+x_2+\cdots + x_n \right) \left(\frac{1}{x_1}+\frac{1}{x_2} + \cdots + \frac{1}{x_n} \right)$$ by Titu's Lemma(a variant of Cauchy Schwarz) $$\Rightarrow \frac{x_1+x_2 + \cdots + x_n}{n} \ge \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}}$$ as desired $\blacksquare$