show that $\frac{c^2}{a}+\frac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$ Let $a,c>0$ show that
$$\dfrac{c^2}{a}+\dfrac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$$
It seem AM-GM inequality.How?

Write $a=b^2$ and $c=d^2$, and multiply through by $b^2d^2$. The inequality becomes
$$b^6+d^6+16b^3d^3\geq 9b^4d^2+9b^2d^4.$$
Dividing by $d^6$ and writing $x=\frac{b}{d}$:
$$x^6+1+16x^3\geq 9x^4+9x^2.$$
This follows from 
$x^6+16x^3+1-9x^4-9x^2=(x-1)^4(x^2+4x+1)\geq 0$.
