Proving Inequalities Involving Positive Real Numbers

[Sam Morse]

Homework Statement

Let a,b,c be positive real numbers. Prove that:

a³+b³+c³≥a²b+b²c+c²a

Homework Equations

The Attempt at a Solution

I assumed that a≥b≥c>0 following which I shifted the left side of this inequality to the right side giving

a³+b³+c³-(a²b+b²c+c²a)≥0

How do I do the required factorisation ... ?

 

[mfb]

Are you sure that there is an easy way to factorize this?

I would expect that this can be reduced to a problem similar to a/b + b/c + c/a > 3.

 

[Infrared]

I see a way to solve it, though it is a bit messy. First divide the entire equation by abc .
A bit of algebra gets $a^3+b^3+c^3 \geq ab + bc+ ac$. We can assume that a,b, and c are all greater than 1 since if (a,b,c) satisfies the inequality, then so does (ax,bx,cx) for any positive x. Using the fact that $x^3 \geq x^2 if x>1$, it is therefore sufficient to show that
$a^2+b^2+c^2 \geq ab + bc +ac$

Get everything over to the left hand side and multiply by 2. You will see that it factors nicely.

Edit: Oops, disregard what I said, I made a mistake in my algebra.

 

[Sam Morse]

Well I have found a factorisation of this algebraic expression on a website. It just reduces the expression to something which easily leads us to believe that the inequality is true. Here's the link :

http://smc.math.ca/crux/v31/n8/public_page492-499.pdf