# 1977_Question_3

The first is trivial by Schur’s Inequality, which states that {a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0} for non-negatives a, b, c and r > 0.

Schur’s inequality can be proven (r=1) by assuming that a \geq b \geq c WLOG.
a(a-b)(a-c)+b(b-a)(b-c) \longrightarrow a(a-b)(a-c)-b(a-b)(b-c)

This simplifies to (a-b)(a(a-c)-b(b-c)), which is always \geq 0 for nonnegative a,b,c.
Finally, c(c-a)(c-b) is positive since both (c-a) and (c-b) are negative.
(For (ii) I’m not so sure as of now)

3 Likes

I keep trying to work out (ii)

I know there’s a way to solve it by expanding it all out, factoring, and brute bashing with the trivial inequality. But that’s a bit ugly.

1 Like

It would be cool to have multiple solutions from people on this topic. Everyone gets credit in the competition when that happens

Here to the rescue for part ii) :

2 Likes

Well done!

1 Like