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

Also, we can use SOS here:
$$\sum_{cyc}\sqrt{a^2+ab+b^2}-\sqrt3(a+b+c)=\sum_{cyc}\left(\sqrt{a^2+ab+b^2}-\frac{\sqrt3(a+b)}{2}\right)=$$
$$=\sum_{cyc}\frac{4(a^2+ab+b^2)-3(a+b)^2}{2\left(2\sqrt{a^2+ab+b^2}+\sqrt3(a+b)\right)}=\sum_{cyc}\frac{(a-b)^2}{2\left(2\sqrt{a^2+ab+b^2}+\sqrt3(a+b)\right)}\geq0.$$
