Iran 1996 Inequality

Problem: (Iran 1996) Prove that \forall x,y,z\in\mathbb{R}^ + ;
(xy + yz + zx)\left(\sum \frac {1}{(x + y)^2}\right)\geq \frac {9}{4}

Solution ( by Vo Quoc Ba Can):
WLOG, assume x\ge y\ge z. Now we use a lemma:
We claim that \sum_{cyc}\frac{1}{(x+y)^2}\geq \frac{1}{4xy}+\frac{2}{(x+z)(y+z)}
Proof: The lemma is equivalent with:
\left(\frac{1}{x+z}-\frac{1}{y+z}\right)^2\geq \frac{(x-y)^2}{4xy(x+y)^2}
\Longrightarrow 4xy(x+y)^2\geq (x+z)^2(y+z)^2
But, as x\ge y\ge z so 4xy\ge 4y^2\ge (y+z)^2
& (x+y)^2\geq (x+z)^2 so the lemma is proved.
\boxed{\text{Using this lemma: }}
The inequality
\implies (xy+yz+zx)\left(\frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\right)\geq \frac{9}{4}
\implies \frac{1}{4}+\frac{z(x+y)}{4xy}+\frac{2(xy+yz+zx)}{(x+z)(y+z)}\geq \frac{9}{4}
\implies \frac{z(x+y)}{4xy}+2-\frac{2z^2}{(x+z)(y+z)}\geq  2\left[\because 2(xy+yz+zx)=2(x+z)(y+z)-2z^2\right]
\implies \frac{z(x+y)}{4xy}\geq \frac{2z^2}{(x+z)(y+z)}
\implies (x+y)(y+z)(z+x)\geq 8xyz.
This easily follows from AM-GM. Hence proved.
This solution is really genial for its simplicity and brilliancy…. and also the easiest solution (in my opinion) possible. :) :o


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