<s>[INST] 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)?
 [/INST] We need to prove that
$$\sum_{cyc}(3a-a^2+ab)\geq\sum_{cyc}(a+2\sqrt{ab})$$ or
$$2\sum_{cyc}(a-\sqrt{ab})\geq\sum_{cyc}(a^2-ab)$$ or
$$2\sum_{cyc}\left(\sqrt{a}-\sqrt{b}\right)^2\geq\sum_{cyc}(a-b)^2$$ or
$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\left(2-(\sqrt{a}+\sqrt{b})^2\right)\geq0,$$
for which it's enough to prove that
$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\left(2(a+b+c)-(\sqrt{a}+\sqrt{b})^2\right)\geq0$$ or
$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\left((\sqrt{a}-\sqrt{b})^2+2c\right)\geq0,$$
which is obvious.
Done!
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