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?

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.
