If $a+b+c\le 1$ then $3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$

Since $a+b \leqslant 1 $ by Cauchy - Schwarz we get $$\sqrt{a} +\sqrt{b} \leq \sqrt{2}\cdot \sqrt{a+b} \leqslant \sqrt{2} $$ hence $$\frac{(\sqrt{a} +\sqrt{b})^2}{2} \leqslant 1$$ multiplying both sides of the above inequality by $(\sqrt{a} -\sqrt{b} )^2 $ we get $$\frac{(a-b)^2 }{2} \leqslant (\sqrt{a} -\sqrt{b})^2 . $$

Now we have

\begin{align} 3 (a+b+c ) -(\sqrt{a} +\sqrt{b} +\sqrt{c})^2 &= 2a +2b +2c -2\sqrt{ab} -2\sqrt{ac} -2\sqrt{bc} \\ &= (\sqrt{a} -\sqrt{b} )^2 +(\sqrt{a} -\sqrt{c} )^2 +(\sqrt{c} -\sqrt{b} )^2\\ &\geqslant \frac{1}{2} \cdot \left ( (a-b)^2 + (a-c)^2 +(c-b)^2 \right)\\ &= a^2 +b^2 +c^2 -ab -ac -ab \end{align}

what completes the proof.