Prove the inequality a^3/b+b^3/c

[Guest]

Hi, I need to prove something like that:

For a,b,c>0 prove:

[tex]\frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} \geq ab + bc + ca[/tex]

I tried to prove this inequality in many ways, but I simply don't khow to do it correctly. Can anyone help me?

[Guest]

It follows from two applications of the rearrangement inequality
https://en.wikipedia.org/wiki/Rearrangement_inequality
Without loss of generality assume [tex]a\leq b\leq c[/tex], this implies
[tex]a^3\leq b^3\leq c^3[/tex] and [tex]1/c\leq 1/b \leq 1/a[/tex]
By the rearrangement inequality we know that
[tex]a^3\times (1/b)+b^3\times(1/c)+c^3\times(1/a)\geq a^3\times (1/a)+b^3\times(1/b)+c^3\times(1/c)[/tex]

Also the rearrangement inequality tells us that
[tex]a\times a+b\times b+c\times c\geq a\times b+b\times c+c\times a[/tex]

Combining these two inequalities gives the desired result.

Hope this helped,

R. Baber.