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?

Certainly an Overkill
$$\frac{c^2}{a}+\frac{a^2}{c}+16\sqrt{ac} \geq 9(a+c)$$
$$\Longleftrightarrow \left(\frac{c^2}{a}-a\right)+\left(\frac{a^2}{c}-c\right)- 8(\sqrt{c}-\sqrt{a})^2 \geq 0$$
$$\Longleftrightarrow (\sqrt{c}-\sqrt{a})^2 \left[(\sqrt{c}+\sqrt{a})^2 (c+a) -8ac \right] \geq 0$$
$$\Longleftrightarrow (\sqrt{c}+\sqrt{a})^2 (c+a) -8ac  \geq 0$$
$$\Longleftrightarrow (a+c)^2+2(a+c)\sqrt{ac}-8ac \geq 0 \tag{*}$$
By AM-GM Inequality
$$(a+c)^2 \geq 4ac \tag{1}$$
$$2(a+c)\sqrt{ac} \geq 4ac \tag{2}$$
Thus, we see that $(*)$ follows from $(1)$ and $(2)$
