# 2007 Indonesia MO Problems/Problem 3

## Problem

Let $a,b,c$ be positive real numbers which satisfy $5(a^2+b^2+c^2)<6(ab+bc+ca)$. Prove that these three inequalities hold: $a+b>c$, $b+c>a$, $c+a>b$.

## Solution (credit to vedran6)

Assume that $a+b \le c$, so $c-a-b \ge 0$. Let $x = c-a-b$, making $x \ge 0$. Therefore, $c = a+b+x$.

By substitution, we have \begin{align*} 5(a^2 + b^2 + (a+b+x)^2) &< 6(ab + (a+b)(a+b+x)) \\ 5(a^2 + b^2 + a^2 + b^2 + x^2 + 2ab + 2ax + 2bx) &< 6(ab + a^2 + ab + ax + ab + b^2 + bx) \\ 5(2a^2 + 2b^2 + x + 2ab + 2ax + 2bx) &< 6(3ab + a^2 + ax + b^2 + bx) \\ 10a^2 + 10b^2 + 5x + 10ab + 10ax + 10bx &< 18ab+6a^2+6ax+6b^2+6bx \\ 4a^2 + 4b^2 + 5x - 8ab + 4ax + 4bx &< 0 \\ 4(a-b)^2 + x(5+4a+4b) &< 0 \end{align*} Note that because $x, a, b \ge 0$, we must have $x(5+4a+4b) \ge 0$. Additionally, by the Trivial Inequality, $4(a-b)^2 \ge 0$. Thus, we must have $4(a-b)^2 + x(5+4a+4b) \ge 0$, is a contradiction. Therefore, we must have $a+b > c$.

Because of symmetry in the equation, we can use similar steps to prove that $b+c>a$ and that $c+a>b$.