<s>[INST] 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:)
 [/INST] Here is a very short one.
Suppose w.l.o.g. $a \ge b \ge c$. Then by the rearrangement inequality,
$$
S = \frac{1}{(a+b)b} +\frac{1}{(b+c)c} + \frac{1}{(a+c)a} \ge \frac{1}{(a+b)c} +\frac{1}{(b+c)a} + \frac{1}{(a+c)b}  = T
$$
So
$$
2 S \ge S + T = \frac{b+c}{(a+b)bc} +\frac{c+a}{(b+c)ca} + \frac{a+b}{(a+c)ab} \geq 3 \Big( \frac{1}{abc} \Big)^{2/3} = 3
$$
which proves it, where in the last step AM-GM was used.
</s>