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)?
 A: 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.
A: We note that
$$
\text{LHS}=\sqrt{(-(b/2)+a)^2+(\sqrt{3}b/2)^2} + \sqrt{((c/2)-a)^2+(\sqrt{3}c/2)^2}\tag{$*$}
$$
so the Minkowski inequality applies to give
$$
\text{LHS}\geq\sqrt{\frac{1}{4}(c-b)^2+\frac{3}{4}(b+c)^2}=\text{RHS}.
$$
Equality is if and only if there is some $\lambda\geq 0$ such that
$$
(a-b/2,\sqrt{3}b/2)=\lambda(-a+c/2,\sqrt{3}c/2).
$$

p.s.: The version of the Minkowski inequality that we have used is
$$
(|x_1|^p+|x_2|^p)^{1/p}+(|y_1|^p+|y_2|^p)^{1/p}\geq(|x_1+y_1|^p+|x_2+y_2|^p)^{1/p}
$$
with $p=2$ and $x_1,x_2,y_1$ and $y_2$ are as in ($*$) (in that order). (e.g. $x_1=a-b/2$, etc.)
