<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] 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.
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