We write $a=1-b-c,b=1-c-a,c=1-a-b$ and replace the $a,b,c$ in the denominators. We get $\sum\sqrt{\frac{ab}{(1-a)(1-b)}}\le\frac 32$. That's equivalent to $\sum\sqrt{\frac{ab}{(a+c)(b+c)}}\le\frac 32$. We now apply AM-GM to each of the square roots in the form $2\sqrt{\frac{ab}{(a+c)(b+c)}}\le\frac a{a+c}+\frac b{b+c}$ and add up the three inequalities. The LHS will be twice our expression, and the RHS will be exactly $3$.

ooh, nice.