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.

Lemma For all $x$ we have $$\sqrt{x^2+x+1}\geq {\sqrt{3}\over 2}(x+1)$$
Proof After squaring and clearing the denominator we get $$4x^2+4x+4\geq 3(x^2+2x+1)$$
which is the same as $$x^2-2x+1\geq 0$$

Using lemma we get  $$\sqrt{a^2+ab+b^2}= b\sqrt{\Big({a\over b}\Big)^2+{a\over b}+1}\geq b\cdot  {\sqrt{3}\over 2}({a\over b}+1)=$$ $$={\sqrt{3}\over 2}(a+b) $$ so
$$... \geq  {\sqrt{3}\over 2}(a+b) + {\sqrt{3}\over 2}(b+c)+  {\sqrt{3}\over 2}(c+a) =3\sqrt{3}$$
