Mathematical Reflections 2011, Issue 6 - Problem S213

Problem:
Let $a,b,c$ be positive real numbers such that $a^2 \geq b^2+bc+c^2$. Prove that
$$a > \min(b,c) + \dfrac{|b^2-c^2|}{a}.$$

Proposed by Titu Andreescu.

Solution:
Without loss of generality, suppose that $\min(b,c)=c$. If $a \geq b+c$, then
$$a^2-ac \geq ab > (b+c)(b-c)=b^2-c^2.$$ If $a<b+c$, then
$$a^2-ac > a^2-(b+c)c \geq b^2 > b^2-c^2.$$ So, for any $a,b,c$ satisfying the required conditions, we have $a > \min(b,c) + \dfrac{|b^2-c^2|}{a}$.