If $a+b+c\le 1$ then $3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$ 
Let $a,b,c\ge 0$ such that $a+b+c\le 1$, prove that
  $$3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge  (\sqrt{a}+\sqrt{b}+\sqrt{c})^2\tag{1}$$

I conjecture:
Let $a_{i}\ge 0$, $i=1,2,\cdots$, $a_{1}+a_{2}+\cdots+a_{n}\le 1$, $n\ge 3$,then 
$$n(a_{1}+a_{2}+\cdots+a_{n})-(a^2_{1}+a^2_{2}+\cdots+a^2_{n}-a_{1}a_{2}-a_{2}a_{3}-\cdots-a_{n}a_{1})\ge (\sqrt{a_{1}}+\sqrt{a_{2}}+\cdots+\sqrt{a_{n}})^2$$
This inequality is stronger than Cauchy-Schwarz inequality,because
$$(a^2+b^2+c^2-ab-bc-ac)\ge 0$$
Applying Cauchy-Schwarz inequality 
$$3(a+b+c)\ge(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$
How to prove the required statement (1)?

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.
