# Problem with factorisation

1. Mar 23, 2013

### Sam Morse

1. The problem statement, all variables and given/known data

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

a3+b3+c3≥a2b+b2c+c2a

2. Relevant equations

3. 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 ... ???

2. Mar 23, 2013

### Staff: Mentor

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.

3. Mar 23, 2013

### 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.

Last edited: Mar 23, 2013
4. Mar 23, 2013