Prove $(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x})^3+12\ge 13(x^3+y^3+z^3)$ for positives $xyz = 1$

Question

let $x,y,z>0$, and such $xyz=1$,show that $$\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\right)^3+12\ge 13(x^3+y^3+z^3)\tag{1}$$

I have konwn use C-S we have $$\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\right)\ge\dfrac{(x+y+z)^2}{x+y+z}=x+y+z$$ this relsut can't sove this inequality $(1)$,so How to prove it?and if use $pqr$ methods,$x+y+z=p,xy+yz+xz=q,xyz=r=1$,but $$\sum \dfrac{x^2}{y}=\dfrac{\sum x^3z}{xyz}=\sum x^3z$$

Answer 1

The hint.

By the $uvw$'s technique prove that: $$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq\frac{5(x^2+y^2+z^2)-2(xy+xz+yz)}{x+y+z}.$$ After this by $uvw$ prove that: $$\frac{5(x^2+y^2+z^2)-2(xy+xz+yz)}{x+y+z}\geq\sqrt[3]{13(x^3+y^3+z^3)-12xyz}.$$ A proof of the second inequality.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$, $xyz=w^3$ and $u^2=tv^2$.

Thus, $t\geq1$ and we need to prove that $$\frac{5(9u^2-6v^2)-6v^2}{3u}\geq\sqrt[3]{13(27u^3-27uv^2+3w^3)-12w^3}$$ or $$(5u^2-4)^3\geq u^3(13u^3-13uv^2+w^3).$$ Now, since $uw^3\leq v^4,$ it's enough to prove that $$(5t-4)^3\geq t(13t^2-13t+1)$$ or $$(t-1)(112t^2-175t+64)\geq0$$ or $$t-1+7(t-1)^2(16t-9)\geq0,$$ which is obvious for $t\geq1$.

Answer 2

Supplement to Michael Rozenberg's nice answer:

We give an alternative proof of the following result.

Fact 1: Let $x, y, z > 0$. Then $$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq\frac{5(x^2+y^2+z^2)-2(xy+xz+yz)}{x+y+z}.$$

We use the SOS-Schur method. Please see e.g. this question for some proofs using SOS-Schur method. In Andreas's nice answer, Remark 2, he gave an algorithm to find $M, N$. That algorithm gives the same $M, N$ as mine.

Proof of Fact 1:

WLOG, assume that $z = \max(x, y, z)$.

We have \begin{align*} &\left[\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}-\frac{5(x^2+y^2+z^2)-2(xy+xz+yz)}{x+y+z}\right]xyz(x + y + z)\\[6pt] ={}& M(x - y)^2 + N(z - x)(z - y) \end{align*} where \begin{align*} M &= x^2(2z - y) + y^2(2z - x),\\ N &= y(y + z - 2x)^2 + x(x - 2y)^2 + y^3 + xy (z - x). \end{align*}

Clearly, $M, N \ge 0$.

We are done.