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.

By $a + b + c = 3$, the inequality is equivalent to
$$
\sum_{cyc}{\sqrt{(3-c)^2-ab}}\geqslant \sqrt{3}
$$
$$
\sum_{cyc}{9 - 6c + c^2 - ab}\geqslant 3
$$
Which is true because $\sum_{cyc}9-6c = 9$ and $a^2 + b^2 + c^2\geqslant ab + bc + ca$.
And $\sum_{cyc}{\sqrt{((3-c)^2-ab)((3-b)^2 - ca)}} \geqslant0$ is true.
