# Proof for Inequality

## Homework Statement

If a, b and c are distinct positive numbers, show that
$2 (a^3 + b^3 + c^3) > a^2b + a^2c + b^2c + b^2a + c^2a + c^2b$

## The Attempt at a Solution

I have tried to expand from $$(a+b+c)^3 > 0$$, also tried $$(a+b)^3 + (b+c)^3 + (c+a)^3 > 0$$, and then $$\frac{a+b+c}{3} > \sqrt[3]{abc}$$. But with no avail. I guess I'm heading in the wrong direction?

Last edited:

Related Precalculus Mathematics Homework Help News on Phys.org
hunt_mat
Homework Helper
Try playing with:
$$a(a-b)^{2}+b(b-c)^{2}+c(c-a)^{2}>0\quad a(a-c)^{2}+b(b-a)^{2}+c(c-b)^{2}>0$$
along with other things like this to get the answer.

Hmm, thanks for your tip off, but I still can't seem to make it...

The 2 is always appearing on the right side of the inequality, but as of the question it's on the right side, if I expand $$(a-b)^2$$ the 2 sticks together with ab instead of $$a^3$$...

disregardthat
Have you heard of the arithmetic mean and geometric mean and their relationship? Try using the AM-GM inequality (it is extremely useful in situations like this, look it up!). Rearrange the left hand side as such: $$\frac{a^3+a^3+b^3}{3} +...$$ Apply the AM-GM inequality at each term and note exactly when you have equality. An alternative way is to use the rearrangement inequality. Assuming without loss of generality a >= b >= c, the inequality follows immediately, and strict inequality follows from a>c.
Have you heard of the arithmetic mean and geometric mean and their relationship? Try using the AM-GM inequality (it is extremely useful in situations like this, look it up!). Rearrange the left hand side as such: $$\frac{a^3+a^3+b^3}{3} +...$$ Apply the AM-GM inequality at each term and note exactly when you have equality.
I think I got it, by using $$\frac{a+b}{2}>\sqrt{ab}$$, etc and by multiplying a and b respectively then adding up all together made it. Thanks for all the tips!