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.

I’m not 100% sure but this is what I got.
$$
a^2+ab+b^2=(a+b/2)^2+\frac{3}{4}b^2>(a+b/2)^2
$$
From this we get
$$
a+b/2+b+c/2+c+a/2=9/2>\sqrt3
$$
Or
$$
a^2+ab+b^2=(a+b/2)^2+\frac{3}{4}b^2> \frac{3}{4}b^2
$$
From this we get
$$
\frac{\sqrt{3}}{2}(a+b+c)= \frac{3\sqrt{3}}{2}>\sqrt{3}
$$
