A inequality proposed at Zhautykov Olympiad 2008 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:)

Let $a=\frac{x}{y}$ and $b=\frac{y}{z}$, where $x$, $y$ and $z$ be positives.
Hence, since $abc=1$, we get $c=\frac{z}{x}$ and by C-S we obtain:
$$\sum_{cyc}\frac{1}{(a+b)b}=\sum_{cyc}\frac{z^2}{xz+y^2}=\sum_{cyc}\frac{z^4}{xz^3+z^2y^2}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(x^2y^2+x^3y)}.$$
Thus, it remains to prove that
$$2(x^2+y^2+z^2)^2\geq3\sum\limits_{cyc}(x^2y^2+x^3y)$$ or
$$\sum_{cyc}(2x^4-3x^3y+x^2y^2)\geq0$$ or
$$\sum_{cyc}(x-y)x^2(2x-y)\geq0$$ or
$$\sum_{cyc}\left((x-y)x^2(2x-y)-\frac{x^4-y^4}{4}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2(7x^2+2xy+y^2)\geq0.$$
Done!
