<s>[INST] How to solve this inequality : Do you know what is this problem or how to solve it ?
Let $x,y,z\in\mathbb{R}^+$ such that $x+y+z=1$. Proove that 
$$\frac{(xy+yz+zx+1)(3x^3+3y^3+3z^3+1)}{9(x+y)(y+z)(z+x)}\ge\left(\frac{x\sqrt{x+1}}{\sqrt[4]{9x^2+3}}+\frac{y\sqrt{y+1}}{\sqrt[4]{9y^2+3}}+\frac{z\sqrt{z+1}}{\sqrt[4]{9z^2+3}}\right)^2.$$
My attempts:
Let $a=\sqrt[4]{x^2+3}$, $b=\sqrt[4]{y^2+3}$, $c=\sqrt[4]{z^2+3}$ but it seems useless 
Thanks a lot  !
 [/INST] If I understood right you used homogenization and assumption $x+y+z=3$.
Now use C-S:
$$\sqrt[4]{x^2+3}=\sqrt{\frac{1}{2}\sqrt{(1+3)(x^2+3)}}\geq\sqrt{\frac{1}{2}(x+3)}.$$
Also, use Holder: $$9(x^3+y^3+z^3)\geq(x+y+z)^3.$$
After this you'll get something obvious:
$$\sum_{cyc}(x-y)^2(x+y)\geq0.$$ 
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