# 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

Titu's Lemma: The sum of multiple fractions in the form $\frac{a_n^2}{b_n}$ where $a_n$ and $b_n$ are sequences of real numbers is greater than of equal to the square of the sum of all $a_i$ divided by the sum of all $b_i$, where i is a whole number less than n+1. Titu's Lemma can be proved using the Cauchy-Schwarz Inequality after multiplying out the denominator of the RHS.

Consider the LHS of the proposed inequality. Split up the numerators and multiply both sides of the fraction by either a or 3a to make the LHS $$\sum_{cyc} \frac {a^4} {5a^2+ab}+\sum_{cyc} \frac {9a^4} {15ac+3a^2}$$ (Cyclic notation is a special notation in which all permutations of (a,b,c) is the summation command.)

Then use Titu's Lemma on all terms: $$\sum_{cyc} \frac {a^4} {5a^2+ab} +\sum_{cyc} \frac {9a^4} {15ac+3a^2} \ge \frac {16(a^2+b^2+c^2)^2} {8(a^2+b^2+c^2)+16(ab+bc+ca)} \ge \frac {16(a^2 + b^2 + c^2)^2}{24(a^2 + b^2 + c^2)} = \frac{2}{3} (a^2 + b^2 + c^2)$$ owing to the fact that $a^2+b^2+c^2 \ge ab+bc+ca$, which is actually equivalent to $(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0$!

## 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)

## Solution 4

By Cauchy-Schwarz,

$$\left(\sum_{cyc} \dfrac{a^3}{5a + b} + \dfrac{b^3}{5a + b} + \dfrac{b^3}{5a + b} + \dfrac{b^3}{5a + b} \right) \ge \dfrac{\left( \sum_{cyc} a^2 + b^2 + b^2 + b^2 \right)^2}{ \left( \sum_{cyc} a(5a + b) + b(5a + b) + b(5a + b) + b(5a + b) \right) }$$

$$= \dfrac{\left(4a^2 + 4b^2 + 4c^2\right)^2}{\left(8a^2 + 8b^2 + 8c^2\right) + \left(16ab + 16bc + 16ca\right)}$$

$$\ge \dfrac{16\left(a^2 + b^2 + c^2\right)^2}{\left(8a^2 + 8b^2 + 8c^2\right) + \left(8a^2 + 8b^2\right) + \left(8b^2 + 8c^2\right) + \left(8c^2 + 8a^2\right)}$$ (by AM-GM) $$= \dfrac{2}{3}\left(a^2 + b^2 + c^2\right)$$ as desired.