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.

Use QM-AM to obtain \begin{align*}\sum_{cyc}\sqrt{a^2+ab+b^2}\geqslant \sum_{cyc}\frac{a+b+\sqrt{ab}}{\sqrt{3}}&=\frac{2(a+b+c)}{\sqrt{3}}+\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{\sqrt{3}}\\&=2\sqrt{3}+\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{\sqrt{3}}\\&>\sqrt{3}\end{align*}
