show this inequality $\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)^3+12\ge 13(x^3+y^3+z^3)$ 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
$$\sum \dfrac{x^2}{y}=\dfrac{\sum x^3z}{xyz}=\sum x^3z$$

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 $t\geq1$.
