<s>[INST] $\sqrt{ab}$ + $\sqrt{bc}$ + $\sqrt{ac}$ $\leq$ $ \frac{3}{2}$ 
If  $ a, b, c \in R^{+}$ and  $\frac{1}{a + 1}$ + $\frac{1}{b + 1}$ + $\frac{1}{c + 1}$ = 2,  then $\sqrt{ab}$ + $\sqrt{bc}$ + $\sqrt{ac}$ $\leq$  $ \frac{3}{2}$.  

I  don't  know  how  to  start.
 [/INST] The condition gives $\sum\limits_{cyc}\left(\frac{1}{1+a}-1\right)=-1$ or $\sum\limits_{cyc}\frac{a}{1+a}=1$.
Thus, by C-S $1=\sum\limits_{cyc}\frac{a}{1+a}\geq\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{3+a+b+c}$, which gives what you wanted.
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