Show $\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$ for $a,b,c>0$ If $a$, $b$, and $c$ are positive real numbers, show that
$$
\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \geq \sqrt{b^2 + bc + c^2}.
$$
What values of $a$, $b$, and $c$ ( or relation among them)  make both sides equal ( makes it an equality)?

Consider the points $A\equiv (0,0),B \equiv (a,0), C,D$ on the plane such that $AC=b$ and $\angle CAB=60^{\circ}$ and $AD=c$ and $\angle DAB=60^{\circ}$. Then $BC=\sqrt{a^2-ab+b^2}$ and $BD=\sqrt{a^2-ac+c^2}$. Now we also have $CD=\sqrt{b^2+bc+c^2}$. Consider the triangle $BCD$ and apply triangle inequality to get $BC+BD\geq CD$ which is the given inequality. For equality we need $B\in CD$. Finding the exact condition should be easy now.
