An inequality proposed at Zhautykov Olympiad 2008.
Let be $a,b,c >0$ with $abc=1$. Prove that:
$$\sum_{\mathrm{cyc}}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$
Set $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.
Our inequality becomes:
$$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \frac{3}{2}.$$
Now we use that: $z^2+x^2 \geq 2zx.$
$$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \sum_{\mathrm{cyc}}{\frac{2z^2}{z^2+x^2+2y^2}} \geq \frac{3}{2}.$$
Now applying Cauchy-Schwarz we obtain the desired result.
What I wrote can be found on this link: mateforum.
But now, I don't know how to apply Cauchy-Schwarz.
Thanks:)
Best Answer
Since $\mathrm{LHS}$ of last inequality is homogeneous we can assume $x^2 + y^2 + z^2 = 1$. Then it becomes $$ \mathrm{LHS} = 2\sum_{cyc} \frac {x^2} {1 + z^2} =:2I $$ Now using Cauchy-Schwarz inequality we get $$ 1 = (x^2 + y^2 + z^2)^2 = \left(\sum_{cyc} x\sqrt{1 + z^2} \cdot \frac x {\sqrt{1 + z^2}}\right)^2 \leq\\ \left(\sum_{cyc} x^2(1 + z^2)\right) \cdot \left( \sum_{cyc} \frac {x^2}{1 + z^2} \right) = I \cdot \sum_{cyc} x^2(1 + z^2) $$ To finish, let's note that CS inequality implies $$ x^2\cdot z^2 + y^2 \cdot x^2 + z^2 \cdot y^2 \leq x^4 + y^4 + z^4 $$ and therefore $$ \sum_{cyc} x^2(1 + z^2) = 1 + x^2 z^2 + y^2 x^2 + z^2 y^2 \leq 1 + \frac {(x^2 + y^2 + z^2)^2} 3 = \frac 4 3 $$