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: $$\text{L.H.S}= a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}$$
$$\le a^2+b^2+c^2+2(ab+bc+ca)$$
$$=(a+b+c)^2=1$$
 A: 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!
A: For $c\rightarrow0^+$ and $a=b=\frac{1}{2}$ we obtain a value $\frac{1}{2}.$
We'll prove that it's an infimum.
Indeed, we need to prove that
$$a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}\geq\frac{1}{2}(a+b+c)^2.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$9u^2-6v^2+6\sqrt{uw^3}\geq\frac{9}{2}u^2$$ and we see that it's enough to prove out inequality for a minimal value of $w^3$, 
which happens in the following cases.


*

*$w^3\rightarrow0^+$.


Let $c\rightarrow0^+$.
In this case we need to prove that
$$a^2+b^2\geq\frac{1}{2}(a+b)^2$$, which is true by C-S:
$$a^2+b^2=\frac{1}{2}(1+1)(a^2+b^2)\geq\frac{1}{2}(a+b)^2;$$
2. Two variables are equal.
Since the inequality $a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}\geq\frac{1}{2}(a+b+c)^2$ is homogeneous and symmetric, we can assume $b=c=1$ and we need to prove that
$$a^2+2+2\sqrt{3a(a+2)}\geq\frac{1}{2}(a+2)^2$$ or
$$4\sqrt{3(a+2)}\geq\sqrt a(4-a),$$ which is obvious for $a>4$, but for $0<a\leq4$ we need to prove that
$$48(a+2)^2\geq a(4-a)^2$$ or $$a^3-8a^2-32a-96\leq0$$ or
$$a^3-64-(8a^2+32a+32)\leq0,$$ which is obviously true.
Done!
The maximal value is indeed, $1$.
For which we need to prove that
$$a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}\leq(a+b+c)^2$$ or
$$ab+ac+bc\geq\sqrt{3abc(a+b+c)}$$ or after squaring of the both sides
$$\sum_{cyc}c^2(a-b)^2\geq0,$$ which is obvious.
A: Let $\psi(x)$ is symmetric function and have gradient on every point of its domain. If $x_i^* = x_j^*$ for any $i,j \in [1,n]$, then $\psi_{x_1}(x^*) = \psi_{x_2}(x^*) = \cdots = \psi_{x_n}(x^*)$. 
Let $\xi_i = (x_1,\cdots,x_{i-1},x_{i+1},\cdots,x_n)$. if $x_i = \varphi(\xi_i)$ and $\nabla x_i = \pm \mathbf{1}$, then there is a Lagrange function $$L(\xi_i,\lambda) = \psi(\xi_i) + \lambda(\alpha - x_i)$$ and $$\nabla L = 0 \longrightarrow \nabla \psi(\xi_i) = \lambda \nabla x_i = \pm \lambda \mathbf{1} $$
where $\psi(\xi_i)$ is also symmetric function.
Therefore, if $\xi_i^* = c\mathbf{1}$ and $x_i = 0$ such that $\varphi(\xi_i^*) = 0$, then $\xi_i^*$ is critical point of $L(\xi_i),$ as well as $\psi(\xi_i)$. 
Hence, $\vartheta_i = (x_1,\cdots,x_{i-1},0,x_{i+1},\cdots,x_n)$ is critical point of $\psi(x)$ for any $i \in [1,n]$.
The symmetric function $f(x) = x_1^2 + x_2^2 + x_3^2 + a\sqrt{x_1,x_2,x_3}$ where $x_1 + x_2 + x_3 = b$ thereby have critical point $$\vartheta_3 = \left(\frac{b}{2},\frac{b}{2},0\right)$$
and $$\min f(x) = f\left(\vartheta_3\right) = \frac{b^2}{2}.$$
For instance $b = 1$, it is $$\min f(x) = \frac{1}{2}.$$
