<s>[INST] $ \frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$ 
For $x,y,z>0,$ I have to prove that
  $$ \frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2  \ge \frac {9}{8}.$$

I tried to use $$ \sqrt {\frac {1}{3} \sum\limits_{cycl}^{} (\frac {x+y}{x+y+ 2 z})^2} \ge \frac {1}{3} \sum\limits_{cycl}^{} \frac {x+y}{x+y+ 2 z} \ge \sqrt[3] {(\frac {x+y}{x+y+ 2 z}) (\frac {y+z}{2 x +y +z}) (\frac {z + x}{z+2 y+z})},$$
but it looks hard to continue and $ \frac {3 x y z}{(x+y) (y+z) (z+x)}$ seems to be not helpful. Any help? Thank you.
 [/INST] Let $\frac{2x}{y+z}=a$, $\frac{2y}{x+z}=b$ and $\frac{2z}{x+y}=c$.
Thus, $ab+ac+bc+abc=4$ and we need to prove that
$$\sum_{cyc}\frac{1}{(a+1)^2}+\frac{3}{8}abc\geq\frac{9}{8}.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$ and $abc=w^3$.
Hence, the condition does not depend on $u$.
In another hand, by AM-GM $$4=ab+ac+bc+w^3=3v^2+w^3\leq3v^2+v^3,$$
which gives $v\geq1$ and since $u\geq v$, we obtain $u\geq1$.
Also, $$4=3v^2+w^3\geq3w^2+w^3,$$ which gives $w\leq1$.
Now, we need to prove that
$$\frac{\sum\limits_{cyc}(ab+a+b+1)^2}{(abc+ab+ac+bc+a+b+c+1)^2}+\frac{3}{8}w^3\geq0$$ or
$$\frac{\sum\limits_{cyc}(a^2b^2+2a^2+1+2a^2b+2a^2c+4ab+4a)}{(w^3+3v^2+3u+1)^2}+\frac{3}{8}w^3\geq\frac{9}{8}$$ or
$$\frac{9v^4-6uw^3+18u^2-12v^2+3+18uv^2-6w^3+12v^2+12u}{(w^3+3v^2+3u+1)^2}+\frac{3}{8}w^3\geq\frac{9}{8}$$ or $f(u)\geq0,$ where
$$f(u)=\frac{3v^4-2uw^3+6u^2+1+6uv^2-2w^3+4u}{(w^3+3v^2+3u+1)^2}+\frac{1}{8}w^3-\frac{3}{8}.$$
But
$$f'(u)=\tfrac{-2w^3+12u+6v^2+4}{(w^3+3v^2+3u+1)^2}-\tfrac{2\cdot3(3v^4-2uw^3+6u^2+1+6uv^2-2w^3+4u)}{(w^3+3v^2+3u+1)^3}=$$
$$=\frac{2(9uw^3+9uv^2+9v^2+7w^3-w^6-1)}{(w^3+3v^2+3u+1)^3}>0,$$
which says that $f$ increases and since the condition does not depend on $u$, it's enough to prove our inequality for a minimal value of $u$, which happens for an equality case of two variables.
Let $b=a$.
Hence, the condition gives $$a^2+2ac+a^2c=4$$ or
$$ac(2+a)=4-a^2$$ or
$$c=\frac{2-a}{a},$$ where $0<a<2$ and we need to prove that
$$\frac{2}{(a+1)^2}+\frac{1}{\left(\frac{2-a}{a}+1\right)^2}+\frac{3}{8}a^2\cdot\frac{2-a}{a}\geq\frac{9}{8}$$ or
$$(a-1)^2(7+2a-a^2)\geq0,$$ which is obvious.
Done!
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