# work 024 – mathlinks user wya

Problem

Let $a,b,c$ be positive real numbers. Prove that

$\frac {a^{2} + bc}{b^{2} + ca} + \frac {b^{2} + ca}{c^{2} + ab} + \frac {c^{2} + ab}{a^{2} + bc}\le\frac {b}{a} + \frac {c}{b} + \frac {a}{c}.$

Proof

$\sum \frac {a^2 + bc}{b^2 + ca} = \sum \frac {\frac {a^2}{c^2} + \frac {b}{c} }{\frac {b^2}{c^2} + \frac {a}{c}}$
Let $\frac {b}{a} = x , \cdots$
therefore , $xyz = 1$
we have to prove that –
$x + y + z \ge \sum x \frac {z + y}{y + x}$

$\Leftrightarrow \sum x \cdot \frac {(x - z)}{x + y} \ge 0$

$\Leftrightarrow \sum (xy+xz)(x^2-z^2) \ge 0$

$\Leftrightarrow x^3y + y^3z + z^3y \ge xyz(x + y + z)$

$\Leftrightarrow \sum \frac {x^2}{y} \ge x + y + z$
which is true by C-S 🙂

Equality hold for $x=y=z$ .

$G(b,a,c) \ge G(a^2+bc,b^2+ca,c^2+ab)$