<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] 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 $$\min f(x) = \frac{1}{2}.$$
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