A inequality proposed at Zhautykov Olympiad 2008

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 $$ \sum_{cyc} x^2(1 + z^2) = 1 + x^2 z^2 + y^2 x^2 + z^2 y^2 \leq 1 + \frac {(x^2 + y^2 + z^2)^2} 3 = \frac 4 3 $$

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.