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?
 A: 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.
A: An Algebraic Solution:
Note that $$\sqrt{a^2-ab+b^2}=\sqrt{\left(a-\frac{b}{2}\right)^2+\left(\frac{\sqrt{3}b}{2}\right)^2}$$ and $$\sqrt{a^2-ac+c^2}=\sqrt{\left(\frac{c}{2}-a\right)^2+\left(\frac{\sqrt{3}c}{2}\right)^2}\,.$$ Using the Triangle Inequality in the form $\sqrt{\sum_{i=1}^nx_i^2}+\sqrt{\sum_{i=1}^ny_i^2}\geq \sqrt{\sum_{i=1}^n\left(x_i+y_i\right)^2}$ for real numbers $x_i$'s and $y_i$'s, we have $$\sqrt{a^2-ab+b^2}+\sqrt{a^2-ac+c^2}\geq \sqrt{\Bigg(\left(a-\frac{b}{2}\right)+\left(\frac{c}{2}-a\right)\Bigg)^2+\left(\frac{\sqrt{3}b}{2}+\frac{\sqrt{3}c}{2}\right)^2}\,.$$ Expanding the right-hand side, we get $\sqrt{a^2-ab+b^2}+\sqrt{a^2-ac+c^2}\geq\sqrt{b^2+bc+c^2}$.  The equality holds if and only if there exists $\lambda \geq 0$ such that $\lambda\left(a-\frac{b}{2},\frac{\sqrt{3}b}{2}\right)=\left(\frac{c}{2}-a,\frac{\sqrt{3}c}{2}\right)$ or $\left(a-\frac{b}{2},\frac{\sqrt{3}b}{2}\right)=(0,0)$ (equivalently, $(a,b,c)=(0,0,c)$ or $(a,b,c)=\left(\frac{\lambda b}{\lambda+1},b,\lambda b\right)$ for some $\lambda \geq 0$).
Although the geometric solution by thkim1011 is a more elegant solution, it is more complicated if $a$, $b$, or $c$ can be negative (although you can prove geometrically with essentially the same method, you have to deal with many cases).  My algebraic solution is better in this sense.
