<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] The problem has appeared here various times and maybe elsewhere too for sure. The key to this is to observe the relationship between: $(a-b)^2 \ge 0$ and $a^2+ab+b^2 \ge \dfrac{3(a+b)^2}{4}$. They are equivalent. Using the latter but taking the square root first $3$ times for $(a,b), (b,c),(c,a)$ and add up, yielding the result.
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