Let $a$, $b$, and $c$ be positive real numbers. Let $a$, $b$, and $c$ be positive real numbers. Prove that
$$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$
Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?

The expressions $a^2 - ab + b^2$, $a^2 - ac + c^2$, and $b^2 + bc + c^2$ should remind you of the law of cosines. Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be vectors such that $\vec{a}$ and $\vec{b}$ are separated by 60 degrees, and $\vec{a}$ and $\vec{c}$ are separated by 60 degrees. Consider the triangle formed by the three vectors. This triangle has side lengths 
$$\sqrt{a^2 - ab + b^2}, \sqrt{a^2 - ac + c^2}, \sqrt{b^2 + bc + c^2}$$
according to the law of cosines. Then the inequality follows by the triangle inequality. 
Equality occurs when the three vectors are collinear. The conditions for this shouldn't be too hard to find using basic geometric methods.
