<s>[INST] Prove that $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$ Problem. (Nguyễn Quốc Hưng) Let $0\le a,b,c\le 3;ab+bc+ca=3.$ Prove that $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$$
I have one solution but ugly, so I 'd like to find another. I will post my solution in the answer.
 [/INST] $\textbf{Solution.}$ (Khang Nguyen) We have $$(3-a)(3-b)(3-c)\ge 0 \rightarrow abc+9(a+b+c)\le 36$$
Assume that $$(b-1)(c-1)\ge 0 \rightarrow 36\ge 8a+(9+a)(b+c)\Rightarrow b+c\le \dfrac{(36-8a)}{9+a}$$
Therefore
\begin{align*}
\text{LHS}&=\sqrt{2a+b+c+2\sqrt{(a+b)(a+c)}}+\sqrt{b+c}\\&
\le \sqrt{2a+\dfrac{36-8a}{9+a}+2\sqrt{a^2+3}}+\sqrt{\dfrac{36-8a}{9+a}}\\&\le 3+\sqrt{3}.
\end{align*}
Let $\dfrac{36-8a}{9+a}=x^2\, \left(1\le x\le 2\right),$ after simplify it becomes:
$$ \left( {x}^{2}+8 \right)  \left( \sqrt {3}+3-x \right) \ge \sqrt {{x}^{2
 }+8}\sqrt {{x}^{4}+4\sqrt {3}\sqrt {7{x}^{4}-50{x}^{2}+124}-10
 {x}^{2}+72}$$
Or $$\sqrt{x^2+8} \left( \sqrt {3}+3-x \right) \ge \sqrt {{x}^{4}+4\sqrt {3}\sqrt {7{x}^{4}-50{x}^{2}+124}-10
 {x}^{2}+72}$$
Since $x\le 2\rightarrow \sqrt{3}+3-x\ge \sqrt{3}+3-2>0\rightarrow \text{VT}>0.$ By squaring both sides, we need to prove
$$-2 \left( \sqrt {3}+3 \right)  \left( \sqrt {3}{x}^{2}+{x}^{3}-6{x
}^{2}-10\,\sqrt {3}+8\,x+6 \right) \ge 4\sqrt {3}\sqrt {7\,{x}^{4}-50\,
{x}^{2}+124}$$
It's easy to prove $\text{LHS}>0$ and from here the inequality is equivalent to
$$24\left( 2+\sqrt {3} \right)  \left( {x}^{2}+8 \right)  \left( 2
\sqrt {3}x+{x}^{2}+8\sqrt {3}-9x-10 \right)  \left( x-2 \right) 
\left( x-1 \right) \ge 0,$$
One more time, we can check that this inequality is true for all $1 \le x \le 2$ and done!
Equality holds when $(a,b,c)=(3,1,0)$ and permutations.
See also here.
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