If $a,b,c>0$ prove that $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge ab+bc+ca$ 
If $a,b,c>0$ prove that  $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge ab+bc+ca$

Simplifying yields 
$a^4c+b^4a+c^4b \ge ab^2c^2+a^2b^2c+a^2bc^2$
Which readily follows from muirhead. I read some where that all muirhead 'like' inequalities can be proven with AM GM HM basic inequalities. I tried to prove it using AM GM,but failed. Maybe a clever substitution can clear the clouds?? Is this even possible to do it?If yes, would you share it?

Hint: 
$$\frac{a^3}{b} + ab \geq 2a^2 \\ \frac{b^3}{c} + bc \geq 2b^2
 \\ \frac{c^3}{a} + ac \geq 2c^2 \\
$$
Edit: Basically it's AM-GM , note : $$\frac{a^3}{b}+ab \ge2 \sqrt{\frac{a^3}{b}\cdot ab}=2a^2$$
You can take it from here.
