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, 2011

Algebra Difficulty 7.9 National olympiad, round 2 Prove it Balkan Mathematical Olympiad

Let $x$ , $y$ and $z$ be positive real numbers such that $xyz = 3(x + y + z)$ . Show that
$\frac{1}{x^2(y+1)} + \frac{1}{y^2(z+1)} + \frac{1}{z^2(x+1)} \geq \frac{3}{4(x+y+z)}$
and determine the cases of equality.

Solution

The AM-GM inequality and the condition in the statement yield $x + y + z \geq 9$ , so
$\frac{3}{4(x + y + z)} \sum (x + 1) = \frac{3}{4} \left( 1 + \frac{3}{x + y + z} \right) \leq 1.$
Finally, apply the Cauchy-Schwarz inequality and take into account the condition in statement to get
$\left(\sum \frac{1}{x^2(y+1)}\right) \left(\sum (y+1)\right) \geq \left(\sum \frac{1}{x}\right)^2 \geq 3 \sum \frac{1}{xy} = 1.$
The conclusion follows. Clearly, equality holds if and only if $x = y = z = 3$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.