**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. Squaring and multiplying by n^2 gives n(a_1^2 + a_2^2 + ...) = (a_1 + a_2 + ...)^2, which is true by Cauchy's (setting b_n = 1).

ReplyDelete2. I have no idea. I'll get back to you on that. X.x

[...] which is true by Nesbitt’s Inequality. QED. [...]

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