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.

Some thoughts:
WLOG, assume $a\ge b \ge c$.
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\sqrt{a+b} + \sqrt{b + c} + \sqrt{c + a} 
\le \sqrt{ \left( \frac{a + b}{2} + (b+c) + \frac{c+a}{\sqrt{3}}\right)(2 + 1 + \sqrt{3})}.$$
It suffices to prove that
$$\frac{a + b}{2} + (b+c) + \frac{c+a}{\sqrt{3}} \le 3 + \sqrt{3}.$$
This is true. Is there nice solutions?
