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Bổ đề: $\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y},\forall x,y>0$
Ta có:
$S=\frac{ab}{1+c}+\frac{ac}{1+b}+\frac{bc}{1+a}$
$=\frac{ab}{(a+c)+(b+c)}+\frac{ac}{(a+b)+(c+b)}+\frac{bc}{(a+b)+(a+c)}$
$\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)+\frac{ac}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)+\frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)$
$=\frac{1}{4}\left(\frac{ac+bc}{a+b}+\frac{ab+ac}{b+c}+\frac{ab+bc}{a+c}\right)$
$=\frac{1}{4}(a+b+c)=\frac{1}{4}$
Max$S=\frac{1}{4}\Leftrightarrow a=b=c=\frac{1}{3}$
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