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2012 USAJMO Problems/Problem 3

Revision as of 14:31, 7 May 2012 by Lightest (talk | contribs) (→‎Solution 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 \[\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2\]

Indeed, by AP-GP, we have

\[41 (a^3 + b^3+b^3) \ge 41*3 ab^2\]

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:

\[36(a^3 + 23b^3) \ge (5a + b)(-a^2 + 25b^2)\]

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

Permuting the variables, we have all these three inequalities:

\[\frac{a^3 + 3b^3}{5a + b} \ge -\frac{1}{36} a^2 + \frac{25}{36} b^2\]

\[\frac{b^3 + 3c^3}{5b + c} \ge -\frac{1}{36} b^2 + \frac{25}{36} c^2\]

\[\frac{c^3 + 3a^3}{5c + a} \ge -\frac{1}{36} c^2 + \frac{25}{36} a^2\]

The inequality in question is just the sum of these, hence it is proved.

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

See Also

2012 USAJMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
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