<s>[INST] 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
 [/INST] Consider a point $O$ and the lines $OA$, $OB$, $OC$ of lengths $a,b,c$ respectively such that $\angle AOB=\angle BOC= \angle COA=120^°$. Consider the triangle $\Delta ABC$. using cosine rule you can see that the side lengths of that triangle are $\sqrt{a^2+ab+b^2},\sqrt{b^2+bc+c^2},\sqrt{c^2+ca+a^2}$.
By definition $O$ is the Fermat point of $\Delta ABC$. See the properties of Fermat point and solve this geometric inequality.
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