<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] An other proof:
$$\Longleftrightarrow \sum_{cyc}\dfrac{(2a+b+c)^2}{2a^2+(b+c)^2}\le 8$$
since
$$\dfrac{(2a+b+c)^2}{2a^2+(b+c)^2}\le\dfrac{4}{3}\dfrac{4a+b+c}{a+b+c}$$
this is true because
$$\Longleftrightarrow (2a-b-c)^2(5a+b+c)\ge 0$$
so
$$\sum_{cyc}\dfrac{(2a+b+c)^2}{2a^2+(b+c)^2}\le\dfrac{4}{3}\sum_{cyc}\dfrac{4a+b+c}{a+b+c}=8$$
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