Prove the inequality: $(y+z)*x^4+(x+z)*y^4+(x+y)*z^4\le((x+y+z)^5)/12$ for positive reals.

Question

The inequality is homogeneous, but I am not sure if that helps. Also, I was thinking about considering a concave function and apply Jensen or Karamata on LHS.

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.

Answer 2

Hint. Since the inequality is homogeneous, you may assume wlog that $x+y+z=1$. Thus, we want to prove that for positive real numbers $x,y,z$ that satisfy $x+y+z=1$, the following inequality holds $$\frac1{12}\geqslant \sum x^4\cdot (1-x)$$ Define now $f(x)=x^4\cdot (1-x)$, for $x\in[0;1]$ and notice that $f''(x)=4\cdot(3 - 5 x)\cdot x^2$. Hence, $f$ is convex on $\left[0; \frac35\right]$ and concave on $\left[\frac35; 1\right]$. You are left with some case work: you might assume wlog $x\leqslant y\leqslant z$ and have to see what happens if, (case $1$) $x\leqslant y\leqslant z\leqslant \frac35$ or (case $2$) $x\leqslant y\leqslant \frac35\leqslant z$... But this is (more or less) smooth :)