Let $a,b,c>0$ and $a+b+c+2=abc$, show that $$\sqrt{\dfrac{a}{b+1}}+\sqrt{\dfrac{b}{c+1}}+\sqrt{\dfrac{c}{a+1}}\ge\sqrt{6}$$
Let $a=\dfrac{y+z}{x}.b=\dfrac{z+x}{y},c=\dfrac{x+y}{z}$,the inequality becomes $$\sum_{cyc}\sqrt{\dfrac{y^2+yz}{x(x+y+z)}}\ge\sqrt{6}$$ following it seem hard to solve by C-S
Let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y}$, where $x$, $y$ and $z$ are positives.
Hence, the condition gives $c=\frac{x+y}{z}$ and we need to prove that: $$\sum_{cyc}\sqrt{\frac{\frac{x+y}{z}}{\frac{y+z}{x}+1}}\geq\sqrt6$$ or $$\sum_{cyc}\sqrt{\frac{x(x+y)}{z(x+y+z)}}\geq\sqrt6$$ or $$\sum_{cyc}\frac{2x(x+y)}{2\sqrt{3z(x+y)\cdot2x(x+y+z)}}\geq1.$$ By AM-GM $$2\sqrt{3z(x+y)\cdot2x(x+y+z)}\leq3z(x+y)+2x(x+y+z)=2x^2+2xy+5xz+3yz.$$ Hence, it remains to prove that $$\sum_{cyc}\frac{x^2+xy}{2x^2+2xy+5xz+3yz}\geq\frac{1}{2}$$ or $$\sum_{cyc}(12x^4y^2+12x^3y^3+20x^4yz-27x^3y^2z+3x^3z^2y-20x^2y^2z^2)\geq0,$$ which is true by AM-GM and Rearrangement.
Done!
Sadly, not a complete solution, but at least it makes reasonable the $\sqrt{6}$. I think it can be finished, but i don't know how to solve the last part
First, we can write
$$ \sum_{\text{cyc}} \sqrt{\frac{y^2+yz}{x(x+y+z)}}=\sum_{\text{cyc}}\sqrt{\frac{y\sqrt{6}}{(x+y+z)}\cdot \frac{y+z}{x\sqrt{6}}} $$
(Why this way? I think the equallity is only achieved at $x=y=z$, and I want to have the two fractions equal in that case. This way, if I use some theorem, the equallity cases will work). Now, using $GM-HM$ inequallity, we have
$$ \sum_{\text{cyc}}\sqrt{\frac{y\sqrt{6}}{(x+y+z)}\cdot \frac{y+z}{x\sqrt{6}}} \geq \sum_{\text{cyc}} \frac{2}{\frac{y\sqrt{6}}{x+y+z}+\frac{x\sqrt{6}}{y+z}}=2\sqrt{6} \sum_{\text{cyc}} \frac{y(y+z)}{(x+y+z)(y+z)+6xy}$$
The right side doesn't change if we multiply by a constant, so we can impose $x+y+z=1$. This way, we have
$$ 2\sqrt{6} \sum_{\text{cyc}} \frac{y(y+z)}{(x+y+z)(y+z)+6xy} \geq 2\sqrt{6} \sum_{\text{cyc}} \frac{y(1-x)}{1-x+6xy}$$