Square roots with $a,b,c$

Call the expression $P$ , we see that $P=\displaystyle\sum{\sqrt{\frac{ab}{ab-c+2}}}=\displaystyle\sum{\sqrt{\frac{ab}{ab+a+b+1}}}$ By AM-GM we have $a+b\geq 2\sqrt{ab}$ and the same for $b,c$ and $c,a$ $\therefore \displaystyle\sum{\sqrt{\frac{ab}{ab+a+b+1}}}\leq \displaystyle\sum{\sqrt{\frac{ab}{ab+2\sqrt{ab}+1}}}$ $\Leftrightarrow P\leq \displaystyle\sum{\sqrt{\frac{ab}{(\sqrt{ab}+1)^2}}}=\displaystyle\sum{\frac{\sqrt{ab}}{\sqrt{ab}+1}}$ $\Leftrightarrow 3-P\geq \displaystyle\sum{\frac{1}{\sqrt{ab}+1}}$ By Titu's Lemma we get $\displaystyle\sum{\frac{1}{\sqrt{ab}+1}}\geq \frac{9}{\displaystyle\sum{\sqrt{ab}}+3}$ and $\displaystyle\sum{\sqrt{ab}}+3\leq a+b+c+3=4$ $\therefore \displaystyle\sum{\frac{1}{\sqrt{ab}+1}}\geq \frac{9}{\displaystyle\sum{\sqrt{ab}}+3}\geq \frac{9}{4}$ $\therefore 3-P\geq \frac{9}{4}$ $\Leftrightarrow P\leq\frac{3}{4}$ The equality holds when $a=b=c=\frac{1}{3}$

Call the expression is $P$ $P=\sum \sqrt{\frac{ab}{ab+a+b+1}}=\sqrt{\frac{ab}{(a+1)(b+1)}}\leq 3(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1})$

We are going to prove: $\sum \frac{a}{a+1}\leq \frac{3}{4}a$

$4a\leq 3a(a+1)$

$(3a-1)a\geq 0$ always true with $a\geq \frac{1}{3}$ and $a,b,c$ are positive reals

So $P\leq \frac{3}{4}$ and the equality holds when $a=b=c=\frac{1}{3}$