<s>[INST] How to prove the inequality: $\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$ Prove that:
$$\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$$
subject to the constraints: 
$$x,y,z >0$$
and
$$x+y+z=1.$$
 [/INST] if $a+b+c=1,a,b,c>0$
$$\Longleftrightarrow \sum_{cyc}\dfrac{a^2+2a+1}{3a^2-2a+1}\le 8$$
since
$$\dfrac{a^2+2a+1}{3a^2-2a+1}\le 4a+\dfrac{4}{3}$$
this is true becasuse
$$\Longleftrightarrow 3(4a+1)(a-\dfrac{1}{3})^2\ge 0$$
so
$$\sum_{cyc}\dfrac{(a+1)^2}{2a^2+(1-a)^2}\le 4(a+b+c)+4=8$$
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