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

Question

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?

Have a look at Rearrangement Inequality – Quang Hoang Apr 21 '20 at 13:31 — Quang Hoang, Apr 21 '20 at 13:31

Siddhartha · Accepted Answer · 2020-04-21T13:36:55.433

6

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.

edited Apr 21 '20 at 13:36

answered Apr 21 '20 at 13:28

Siddhartha

2,681

1

$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge 2(a^2+b^2+c^2)-ab-bc-ac \ge 2(ab+bc+ca)-ab-bc-ca=ab+bc+ca$ is that it? – Mathematical Curiosity Apr 21 '20 at 13:55
Yes ! that's it. – Siddhartha Apr 21 '20 at 13:56

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

1 Answers1