Sunday, August 16, 2009

KBB2 Problem 4

For $a,b,c$ positive reals, show that:

$\displaystyle \frac{6a}{9a^2+5(a+b+c)^2} + \frac{6b}{9b^2+5(a+b+c)^2} + \frac{6c}{9c^2+5(a+b+c)^2} \leq \frac{1}{a+b+c}$

Solution

Let $x = \frac{3a}{a+b+c},y = \frac{3b}{a+b+c}, z = \frac{3c}{a+b+c}$. Then $x+y+z = 3$ and our inequality is equivalent to

$\frac{6x}{9x^2+5(x+y+z)^2} + \frac{6y}{9y^2+5(x+y+z)^2} + \frac{6z}{9z^2+5(x+y+z)^2} \leq \frac{1}{x+y+z}$

$\displaystyle \frac{2x}{x^2+5} + \frac{2y}{y^2+5} + \frac{2z}{z^2+5} \leq 1$

Now,

$\displaystyle \frac{2x}{x^2+5} \leq \frac{2x+1}{9}$

$\displaystyle \iff (2x+1)(x^2+5) \geq 18x$

The last inequality is equivalent to $(x-1)^2(2x+5) \geq 0$ which is obviously true, or it can also be proved by AM-GM:

$\displaystyle 2x+1 \geq 3\sqrt[3]{x^2}$

$\displaystyle x^2+5 \geq 6\sqrt[6]{x^2}$

multiplying them gives $(2x+1)(x^2+5) \geq 18x$

Now,

$\displaystyle \frac{2x}{x^2+5} \leq \frac{2x+1}{9}$

$\displaystyle \frac{2y}{y^2+5} \leq \frac{2y+1}{9}$

$\displaystyle \frac{2z}{z^2+5} \leq \frac{2z+1}{9}$

So when we add them, we have our desired inequality.