Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$ Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2}  + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$
So, using AM-GM, or just pop out squares under square roots we can show:
$$\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2}  + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}),$$
i.e. we need next to show that $(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge (a + b + c)$, but i don't know how to do it.
Any help appreciated

By Minkowski (triangle inequality) $$\sum_{cyc}\sqrt{a^2+ab+b^2}=\sum_{cyc}\sqrt{\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2}\geq$$
$$\geq\sqrt{\left(\sum_{cyc}\left(a+\frac{b}{2}\right)\right)^2+\frac{3}{4}\left(\sum_{cyc}b\right)^2}=\sqrt3(a+b+c).$$
