# Difference between revisions of "2012 USAJMO Problems/Problem 3"

## Problem

Let $a$, $b$, $c$ be positive real numbers. Prove that $$\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{2}{3} (a^2 + b^2 + c^2).$$

## Solution

By the Cauchy-Schwarz inequality, $$[a(5a + b) + b(5b + c) + c(5c + a)] \left( \frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \right) \ge (a^2 + b^2 + c^2)^2,$$ so $$\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc}.$$ Since $a^2 + b^2 + c^2 \ge ab + ac + bc$, $$\frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc} \ge \frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = \frac{1}{6} (a^2 + b^2 + c^2).$$ Hence, $$\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).$$

Again by the Cauchy-Schwarz inequality, $$[b(5a + b) + c(5b + c) + a(5c + a)] \left( \frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \right) \ge (a^2 + b^2 + c^2)^2,$$ so $$\frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \ge \frac{(a^2 + b^2 + c^2)^2}{a^2 + b^2 + c^2 + 5ab + 5ac + 5bc}.$$ Since $a^2 + b^2 + c^2 \ge ab + ac + bc$, $$\frac{(a^2 + b^2 + c^2)^2}{a^2 + b^2 + c^2 + 5ab + 5ac + 5bc} \ge \frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = \frac{1}{6} (a^2 + b^2 + c^2).$$ Hence, $$\frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).$$

Therefore, $$\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).$$

## Solution 2

Split up the numerators and multiply both sides of the fraction by either a or 3a: $\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2}$ Then use Titu's Lemma: $\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2} \ge \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)}$ It suffices to prove that$\frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} \ge \frac {2} {3} (a^2+b^2+c^2)$ After some simplifying, it reduces to $a^2+b^2+c^2 \ge ab+bc+ca$ which is trivial by the Rearrangement Inequality. -r31415

## Solution 3

We proceed to prove that $$\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2$$

(then the inequality in question is just the cyclic sum of both sides, since $$\sum_{cyc} (-\frac{1}{36} a^2 + \frac{25}{36} b^2) = \frac{24}{36}\sum_{cyc} a^2 = \frac{2}{3} (a^2+b^2+c^2)$$ )

Indeed, by AP-GP, we have

$$41 (a^3 + b^3+b^3) \ge 41 \cdot 3 ab^2$$

and

$$b^3 + a^2b \ge 2 ab^2$$

Summing up, we have

$$41a^3 + 83b^3 + a^2 b \ge 125 ab^2$$

which is equivalent to:

$$36(a^3 + 3b^3) \ge (5a + b)(-a^2 + 25b^2)$$

Dividing $36(5a+b)$ from both sides, the desired inequality is proved.

--Lightest 15:31, 7 May 2012 (EDT)