## Wednesday, November 23, 2005

### To Equal or not to Equal. Topic: Inequalities. Level: AIME/Olympiad.

Problem #1: Prove that, given two positive reals $a$ and $b$, we have $\frac{a+b}{2} \ge \sqrt{ab}$.

Solution: Begin with the trivial inequality ($x^2 \ge 0$) applied on $a-b$. We have

$(a-b)^2 \ge 0$
$a^2-2ab+b^2 \ge 0$
$a^2+2ab+b^2 \ge 4ab$
$(a+b)^2 \ge 4ab$
$a+b \ge 2\sqrt{ab}$
$\frac{a+b}{2} \ge \sqrt{ab}$

as desired. QED.

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Comment: Note that we have just derived the Arithmetic Mean-Geometric Mean Inequality, more commonly referred to as just AM-GM. This can be generalized to $n$ variables by induction, that is, $\frac{a_1+a_2+\cdots+a_n}{n} \ge \sqrt[n]{a_1a_2 \cdots a_n}$.

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Problem #2: Given two sequences of positive reals $\{a_i\}$ and $\{b_i\}$ for $i = 1,2, \ldots, n$, prove that $(a_1^2+a_2^2+\cdots+a_n^2)(b_1^2+b_2^2+\cdots+b_n^2) \ge (a_1b_1+a_2b_2+\cdots+a_nb_n)^2$.

Solution: We shall prove this with vectors (read the previous blog post to get some background information).

Consider the vector $A = (a_1,a_2,\ldots,a_n)$ and the vector $B = (b_1,b_2,\ldots,b_n)$ (in an $n$-space).

We have $|A||B| \ge |A||B|\cos{\theta} = A \cdot B$ since $\cos{\theta} \le 1$.

But recall that $A \cdot B = (a_1,a_2,\ldots,a_n) \cdot (b_1,b_2,\ldots,b_n) = (a_1b_1+a_2b_2+\cdots+a_nb_n)$.

Also, $|A| = \sqrt{a_1^2+a_2^2+\cdots+a_n^2}$ and $|B| = \sqrt{b_1^2+b_2^2+\cdots+b_n^2}$.

Combining them, we get $\sqrt{a_1^2+a_2^2+\cdots++a_n^2} \cdot \sqrt{b_1^2+b_2^2+\cdots+b_n^2} \ge (a_1b_1+a_2b_2+\cdots+a_nb_n)$, from which our result follows directly upon squaring both sides:

$(a_1^2+a_2^2+\cdots+a_n^2)(b_1^2+b_2^2+\cdots+b_n^2) \ge (a_1b_1+a_2b_2+\cdots+a_nb_n)^2$. QED.

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Comment: This is known as Cauchy's Inequality, or the Cauchy-Schwarz Inequality. It is one of the most useful inequalities to know and can be generalized (see Holder's Inequality).

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Problem #3: Given three positive reals $a$, $b$, and $c$, prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$.

Solution: Let's put our newly-learned inequalities in action!

By AM-GM, we have $a^2+b^2 \ge 2ab$. Applying this to the two other pairs and summing them up, we get

$a^2+b^2+c^2 \ge ab+bc+ca$.

Adding $2(ab+bc+ca)$ to both sides, we have

$a^2+b^2+c^2+2(ab+bc+ca) = (a+b+c)^2 \ge 3(ab+bc+ca) \Rightarrow \frac{(a+b+c)^2}{2(ab+bc+ca)} \ge \frac{3}{2}$.

Rewrite the original LHS (left hand side) as $\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}$. By Cauchy, we have

$(ab+bc+bc+ba+ca+cb)\left(\frac{a^2}{ab+bc}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}\right) \ge (a+b+c)^2$

with $a_1^2 = ab+bc$, $b_1^2 = \frac{a^2}{ab+bc}$, and similarly define $a_2, a_3, b_2, b_3$.

So then dividing by $2(ab+bc+ca)$ and combining it with the previous inequality, we find

$\frac{a^2}{ab+bc}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb} \ge \frac{(a+b+c)^2}{2(ab+bc+ca)} \ge \frac{3}{2}$

as desired. QED.

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Comment: This is known as Nesbitt's Inequality, and many of the classical inequalities can be applied to prove it.

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Practice Problem #1: Prove the Root-Mean-Square - Arithmetic Mean Inequality - $\sqrt{\frac{a_1^2+a_2^2+\cdots+a_n^2}{n}} \ge \frac{a_1+a_2+\cdots+a_n}{n}$.

Practice Problem #2: (IMO 1995 #A2) Given positive reals $a, b, c$ such that $abc = 1$, prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)} \ge \frac{3}{2}$.

1. 2. 