<s>[INST] Minimize value of the function $a^2+b^2+c^2+2\sqrt{3abc}$ Let $a,b,c$ be the positive real numbers such that $a+b+c=1$. Find Minimize of $$P=a^2+b^2+c^2+2\sqrt{3abc}$$

WA says that $P$ gets only a local minimum. But i think it must be maximum value of $P$.
Then by AM-GM: $$\text{L.H.S}= a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}$$
$$\le a^2+b^2+c^2+2(ab+bc+ca)$$
$$=(a+b+c)^2=1$$
 [/INST] Also, we can use Shur here.
Indeed, we need to prove that
$$a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}\geq\frac{1}{2}(a+b+c)^2$$ or
$$a^2+b^2+c^2+4\sqrt{3abc(a+b+c)}\geq2(ab+ac+bc).$$
By AM-GM $$4\sqrt{3abc(a+b+c)}\geq4\sqrt{3abc\cdot3\sqrt[3]{abc}}=12\sqrt[3]{a^2b^2c^2}>3\sqrt[3]{a^2b^2c^2}$$ and it's enough to prove that:
$$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}\geq2(ab+ac+bc).$$
Now, let $a^2=x^3$, $b^2=y^3$ and $c^2=z^3$. 
Thus, by Schur we obtain:
$$\sum_{cyc}\left(a^2-2ab+\sqrt[3]{a^2b^2c^2}\right)=\sum_{cyc}(x^3-2\sqrt{x^3y^3}+xyz)=$$
$$=\sum_{cyc}(x^3-x^2y-x^2z+xyz)+\sum_{cyc}xy(\sqrt{x}-\sqrt{y})^2\geq0$$
and we are done!
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