Given $x,y,z>0$ satisfy $x+y+z=1$. Find the value of minimize $$\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}$$
After use C-S and AM-GM i get $\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+1}$, help me
If $x=y=z=\frac{1}{3}$ then we get a value $\frac{1}{4}$.
We'll prove that this is a minimal value.
Indeed, by C-S $$\sum_{cyc}\frac{x^4}{(x^2+y^2)(x+y)}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(x^2+y^2)(x+y)}.$$ Thus, it remains to prove that $$\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(x^2+y^2)(x+y)}\geq\frac{x+y+z}{4}$$ or $$4(x^2+y^2+z^2)^2\geq(x+y+z)\sum_{cyc}(2x^3+x^2y+x^2z)$$ or $$\sum_{cyc}(2x^4-3x^3y-3x^3z+6x^2y^2-2x^2yz)\geq0$$ or $$\sum_{cyc}(2x^4-3x^3y-3x^3z+4x^2y^2)+\sum_{cyc}(2x^2y^2-2x^2yz)\geq0$$ or $$\sum_{cyc}(x^4-3x^3y+4x^2y^2-3xy^3+y^4)+\sum_{cyc}(x^2z^2-2z^2xy+y^2z^2)\geq0$$ or $$\sum_{cyc}(x-y)^2(x^2-xy+y^2)+\sum_{cyc}z^2(x-y)^2\geq0,$$ which is obvious.
Done!