Proving $\sum_{cyc} \sqrt{a^2+ab+b^2}\geq\sqrt 3$ when $a+b+c=3$ Good evening everyone, I want to prove the following:

Let $a,b,c>0$ be real numbers such that $a+b+c=3$. Then $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geq 3\sqrt 3.$$

My attempt: I try noting that $$\sum_{cyc} \sqrt{a^2+ab+b^2}=\sum_{cyc} \sqrt{(a+b)^2-ab}$$
and now I want to apply Cauchy-Schwarz but it is the wrong direction.

Observe that, in virtue of the QM-AM inequality, we obtain $$\sqrt{a^2+ab+b^2}\geqslant \frac{a+\sqrt{ab}+b}{\sqrt3}$$ Similarly $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geqslant \frac{2\cdot(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{\sqrt3}$$ We thus want to prove $$\frac{2\cdot(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{\sqrt3}\geqslant\sqrt3\iff 6+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\geqslant 3$$ whicht is obsviously true
