Cyclic inequality with fractions

Question

Let $x,y,z\geq0$. Show that $x^4(y+z)+y^4(z+x)+z^4(x+y)\leq\frac{1}{12}(x+y+z)^5$

Answer 1

Suppose $x \geqslant y \geqslant z$ and $$f(x,y,z) = x^4(y+z)+y^4(z+x)+z^4(x+y).$$ We have $$\begin{aligned}f(x,y,z)-f(x,y+z,0) & = y^4(z+x)+z^4(x+y)-x(y+z)^4\\ \\& = y^4(z+x)+z^4(x+y)-x(y+z)^4 \\ \\&=-yz\left[3x(y+z)^2+y^2(x-y)+z^2(x-z)\right] \leqslant 0.\end{aligned}.$$ Therefore $f(x,y,z) \leqslant f(x,y+z,0).$ Finally, we need to prove $$f(x,y+z,0) = x^4(y+z)+x(y+z)^4\leqslant \frac{1}{12}(x+y+z)^5.$$ Setting $a = y +z,$ the inequality become $$12ax(a^3+x^3) \leqslant (x+a)^5,$$ or $$(a+x)(a^2-4ax+x^2)^2 \geqslant 0.$$ Which is true. The proof is completed.