Nice problem: Prove that: $ab+bc+ca \ge \sum{\sqrt{a^2+b^2+3}}$ Problem: Let $a,b,c>0:a+b+c=abc.$ Prove that: $$ab+bc+ca\ge \sqrt{a^2+b^2+3}+\sqrt{b^2+c^2+3}+\sqrt{c^2+a^2+3}$$
Please help me give a hint to get a nice proof!
My attempts after squaring both side, it is desired to show: $a^2b^2+b^2c^2+c^2a^2+4(ab+bc+ca)\ge9+2\sum{\sqrt{(a^2+b^2+3)(b^2+c^2+3)}}$
It seems I am lost!

We need to prove that:
$$ab+ac+bc\geq\sum_{cyc}\sqrt{\frac{abc}{a+b+c}\cdot\left(a^2+b^2+\frac{3abc}{a+b+c}\right)}$$ or
$$(ab+ac+bc)(a+b+c)\geq\sum_{cyc}\sqrt{abc((a^2+b^2)(a+b+c)+3abc)}.$$
Now, by AM-GM and C-S
$$\sum_{cyc}\sqrt{(a^2+b^2)(a+b+c)+3abc}=\sum_{cyc}\sqrt{(a^2+b^2)(a+b)+(a^2+b^2+3ab)c}\leq$$
$$\leq\sum_{cyc}\sqrt{(a^2+b^2)(a+b)+\frac{5(a+b)^2c}{4}}=\frac{1}{2}\sum_{cyc}\sqrt{(a+b)(4a^2+4b^2+5(a+b)c)}\leq$$
$$\leq\frac{1}{2}\sqrt{\sum_{cyc}(a+b)\sum_{cyc}(4a^2+4b^2+5(a+b)c)}=\sqrt{(a+b+c)\sum_{cyc}(4a^2+5ab)}$$ and it's enough to prove that:
$$(a+b+c)(ab+ac+bc)^2\geq abc\sum_{cyc}(4a^2+5ab)$$ or
$$\sum_{cyc}c^3(a-b)^2\geq0.$$
