show that $\frac{c^2}{a}+\frac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$

Question

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?

Answer 1

Yes, we can use AM-GM!

Let $\sqrt{\frac{a}{c}}+\sqrt{\frac{c}{a}}=u$.

Hence, we need to prove that $u^3-3u+16\geq9u$ or

$u^3+8+8\geq12u$, which is AM-GM.

Answer 2

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 $x^6+16x^3+1-9x^4-9x^2=(x-1)^4(x^2+4x+1)\geq 0$.

Answer 3

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 $(2)$