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.

Solution 5

We note that if we can prove that there exists a real number $x$ such that $\frac{a^3+3b^3}{5a+b}\geq xa^2+(\frac{2}{3}-x)b^2$ for all positive reals $a$ and $b$ , then we are done since both sides are a cyclic sum of $a,b$ , and $c$ . Note that if we multiply both $a$ and $b$ by a constant $r$ , the left and right sides will both multiply by $r^2$ (since the numerator on the left will multiply by $r^3$ and the denominator will multiply by $r$ . This means that if the inequality is satisfied for an ordered pair $(a,b)$ , then it is also satisfied for $(ar,br)$ for any positive real $r$ . Thus, without loss of generality we can let $b=1$ ,and our inequality becomes $\frac{a^3+3}{5a+1}\geq a^2x+\frac{2}{3}-x.$ Expanding this and rearranging we get $a^3(1-5x)+a^2(-x)+a(5x-\frac{10}{3})+x+\frac{7}{3}\geq0$ . Since $1$ is a root of the left side, we can factor this as $(a-1)(a^2(1-5x)+a(1-6x)-x-\frac{7}{3})\geq0.$ Note that for this to always be nonnegative, we must have $a^2(1-5x)+a(1-6x)-x-\frac{7}{3}$ be positive when $a>1$ and negative when $a<1$ . Thus, it must have a root at $a=1$ , so we plug in $a=1$ is a root and find that we must have $x=-\frac{1}{36}$ . When $x=-\frac{1}{36}$ , the expression factors as $(a-1)^2(\frac{41}{36}a+\frac{83}{36})\geq0$ which obviously is nonnegative for all positive reals $a$ , so we are done.

~john0512, bronzetruck2016

Solution 6

$LHS=\sum_{cyc}\frac{a^3+3b^3}{5a+b}=\sum_{cyc}\frac{a^3}{5a+b}+\sum_{cyc}\frac{3b^3}{5a+b}=\sum_{cyc}\frac{a^4}{5a^2+ab}+\sum_{cyc}\frac{9b^4}{15ab+3b^2}$ which tells that $LHS\geq \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+ab+bc+ac}+\frac{3(a^2+b^2+c^2)}{(a^2+b^2+c^2)+5(ab+bc+ac)}\geq (\frac{1}{2}+\frac{1}{6})(a^2+b^2+c^2)=RHS$ , our proof is done

~bluesoul

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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