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.

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    $\begingroup$ $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge 2(a^2+b^2+c^2)-ab-bc-ac \ge 2(ab+bc+ca)-ab-bc-ca=ab+bc+ca$ is that it? $\endgroup$ – Mathematical Curiosity Apr 21 at 13:55
  • $\begingroup$ Yes ! that's it. $\endgroup$ – Siddhartha Apr 21 at 13:56

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