Problem. Let $a,b,c\ge 0: a+b+c+abc=4.$ Find minimal value of $P$ $$P=\frac{\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}}{\sqrt{ab+bc+ca+6}}.$$
Source: Vo Quoc Ba Can.
My attempt:
Set $$P(a,b,c)=\frac{\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}}{\sqrt{ab+bc+ca+6}}.$$
After some calculating works, I've tried prove $P(1,1,1)$ is the minimal value.
It means that we need to prove $$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\ge\sqrt{2}\cdot\sqrt{ab+bc+ca+6}.$$
By squaring both side, it remains to prove $$a+b+c+3+2\sum_{cyc}\sqrt{(a+1)(b+1)}\ge 2(ab+bc+ca+6),$$or $$2\sum_{cyc}\sqrt{(a+1)(b+1)}\ge 5+abc+2(ab+bc+ca).$$
I was stucked here. Hope to see some ideas to continue the idea.
Also, all answer and comment are welcome. Thank for your attention.
Updated edit: Thank you @138 Aspen. I made a mistake in calculating.
Minimum should be $$P(0,2,2)=\frac{2\sqrt{30}+\sqrt{10}}{10}.$$
It means that we need to prove $$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\ge\frac{2\sqrt{30}+\sqrt{10}}{10}\cdot\sqrt{ab+bc+ca+6},$$which seems ugly. Hope MV (mixing variables) technique works.
Best Answer
Put $M=\left(\frac{2\sqrt{30}+\sqrt{10}}{10}\right)^2$, $x=\sqrt{a+1}\ge 1$, $y=\sqrt{b+1}\ge 1$, and $z=\sqrt{c+1}\ge 1$.
Thus we need to show $(x+y+z)^2\ge M(ab+bc+ca+6)$.
We have $$x^2+y^2+z^2=a+b+c+3,$$ $$x^2y^2z^2=(a+1)(b+1)(c+1)=$$ $$abc+ab+bc+ca+a+b+c+1=ab+bc+ca+5,$$ $$x^2y^2+y^2z^2+z^2x^2=$$ $$(a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)=$$ $$ab+bc+ca+2(a+b+c)+3=$$ $$ab+bc+ca+2(x^2+y^2+z^2)-3.$$
Thus we have to show that
$$(x+y+z)^2\ge M(x^2y^2+y^2z^2+z^2x^2-2(x^2+y^2+z^2)+9)$$
provided $x,y,z\ge 1$ and $$x^2y^2z^2-5=x^2y^2+y^2z^2+z^2x^2-2(x^2+y^2+z^2)+3.$$
Maybe this can be achieved using Lagrange's multipliers.