Prove that $\frac{1}{\sqrt{ab+a+2}}+ \frac{1}{\sqrt{bc+b+2}}+ \frac{1}{\sqrt{ac+c+2}} \leq \frac{3}{2}$ Let $abc=1$ and $a,b,c>0$.
Prove that 
$$\frac {1}{\sqrt {ab+a+2}}+ \frac {1}{\sqrt {bc+b+2}}+ \frac {1}{\sqrt {ac+c+2}} \leq \frac {3}{2}.$$
I guess that $$\frac {1}{\sqrt {ab+a+2}}+ \frac {1}{\sqrt {bc+b+2}}+ \frac {1}{\sqrt {ac+c+2}} = \frac {3}{2}$$ if and only if $a=b=c=1$. I have spent two nights on this but it's just a mess and I still don't know how to apply Cauchy Schwarz or something to write a rigorous proof for this exercise.
 A: $\mathbf {Hint: }$
Let $$\frac{1}{\sqrt{ab+a+2}}+ \frac{1}{\sqrt{bc+b+2}}+ \frac{1}{\sqrt{ac+c+2}}=A$$
Applying C-S: 
$$\left(\frac {1}{ab+a+2}+\frac {1}{bc+b+2}+\frac {1}{ac+c+2}\right)(3) \ge (A)^2$$
Hence $$A\le \sqrt{\left(\frac {1}{ab+a+2}+\frac {1}{bc+b+2}+\frac {1}{ac+c+2}\right)(3)}$$
Now consider by Titu's lemma we have 
$$\frac {1}{ab+1} + \frac {1}{a+1} \ge \frac {4}{ab+a+2}$$
Hence our obtained inequality now has 
$$A\le \sqrt{
\left(
\frac {1}{ab+a+2}+\frac {1}{bc+b+2}+\frac {1}{ac+c+2}
\right)
(3)
}\le 
\sqrt {
\frac{3}{4}
\sum_{cyc} \left(\frac {1}{ab+1} + \frac {1}{a+1}
\right) }$$
But 
$$\sum_{cyc} \left(\frac {1}{ab+1} + \frac {1}{a+1}\right) =\sum_{cyc} \left(\frac {c}{c+1} + \frac {1}{a+1}\right)=3 $$
 Because $abc=1$
Hence 
$$A \le \sqrt {\frac{3}{4} \sum_{cyc} \frac {c}{c+1} + \frac {1}{a+1}}=\sqrt {\frac{9}{4}}=\frac {3}{2}$$
Hope it helped now. 
A: Let $a=\frac{y}{x}$ and $b=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and by C-S we obtain:
$$\sum_{cyc}\frac{1}{\sqrt{ab+a+2}}=\sum_{cyc}\sqrt{\frac{1}{\frac{z}{x}+\frac{y}{x}+2}}=\sum_{cyc}\sqrt{\frac{x}{2x+y+z}}$$
$$\leq\sqrt{\sum_{cyc}1^2\sum_{cyc}\frac{x}{2x+y+z}}=\sqrt{3\sum_{cyc}x\cdot\frac{1}{x+z+x+y}}$$
$$\leq\sqrt{3\sum_{cyc}\frac{x}{(1+1)^2}\left(\frac{1^2}{x+z}+\frac{1^2}{x+y}\right)}=\sqrt{\frac{3}{4}\sum_{cyc}\left(\frac{x}{x+z}+\frac{x}{x+y}\right)}$$
$$=\sqrt{\frac{3}{4}\sum_{cyc}\left(\frac{y}{y+x}+\frac{x}{x+y}\right)}=\frac{3}{2}.$$
