For every $x,y,z \in[2,7] $ and $x+y+z=12$, prove that $x^3+y^3+z^3\leq 316+x^2+y^2+z^2$

Question

I have tried to use the majorization inequality to approach this question.

I have both $x^3$ and $x^2$ is convex in the region $[2,7]$ since there second derivatives are positive on this interval.

After I rewrite the equation, I have $(x^3-x^2)-(y^3-y^2)-(z^3-z^2)\leq 316$

So, I wonder if I can first say something about $g(x)=x^3-x^2$ by the majorization inequality then generalize to y and z cases.

I would appreciate if someone can help.

Answer 1

Solution with use of the majorization inequality.

Wlog $x\geq y\geq z$. Then we have \begin{align}x&\leq 7\\ x+y&\leq 10\\ x+y+z&=12 \end{align} so $(x,y,z) \preceq (7,3,2)$. Since $g$ in convex on $[2,7]$ we have $$g(x)+g(y)+g(z)\leq g(7)+g(3)+g(2) = 316$$

Answer 2

HINT( for another way).

For $(x,y,z)=(2,3,7)$ (and any of the six permutations) we have exactly $x^3+y^3+z^3= 316+x^2+y^2+z^2$.

Fixing $z=7$ we have to prove $x^3+y^3\le 22 + x^2+y^2$ in which the equality is reached at $(x,y)=(2,3)$ so we must take $z\lt7$. An easy calculation more and we are done.