If $a+b+c=3$, then show $\sqrt{a^2 + ab +b^2}+ \sqrt{b^2 + bc +c^2}+\sqrt{c^2 + ac +a^2} \geq \sqrt{3}$ If $a+b+c=3$, and $a,b,c$ are positive real numbers, then show $\sqrt{a^2 + ab +b^2} + \sqrt{b^2 + bc +c^2} +\sqrt{c^2 + ac +a^2}\geq \sqrt{3}$
Normally when I do inequalities I try to first find where equality would be achieved, but in this case I have no idea where to start to find it.
From $a^2 - 2ab +b^2 \geq 0$ we can obviously get $a^2 -ab +b^2\geq 3ab$, and after substituting that, we can change the inequality to showing that $\sqrt{ab} + \sqrt{bc} +\sqrt{ac} \geq 1$, but clearly if a=b=c, then equality is not achieved. I wanted to try to use Cauchy-Schwarz or something of the sort, but those inequalities generally are used to find the upper bound of sums of roots.

It's true even for any reals $a$, $b$ and $c$ such that $a+b+c=3.$
Indeed, by C-S $$\sum_{cyc}\sqrt{a^2+ab+b^2}=\sqrt{\sum_{cyc}\left(2a^2+ab+2\sqrt{(a^2+ab+b^2)(a^2+ac+c^2)}\right)}=$$
$$=\sqrt{\sum_{cyc}\left(2a^2+ab+2\sqrt{\left(\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2\right)\left(\left(a+\frac{c}{2}\right)^2+\frac{3}{4}c^2\right)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\left(2a^2+ab+2\left(\left(a+\frac{b}{2}\right)\left(a+\frac{c}{2}\right)+\frac{3}{4}bc\right)\right)}=$$
$$=\sqrt{\sum_{cyc}(4a^2+5ab)}\geq\sqrt{3(a+b+c)^2}=\sqrt3|a+b+c|=3\sqrt3>\sqrt3.$$
For positive variables we have:
$$\sum_{cyc}\sqrt{a^2+ab+b^2}\geq a+b+c=3>\sqrt3.$$
