Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
$\frac{ab}{c+1}=\frac{ab}{c+a+b+c}=ab.\frac{1}{(a+c)+(b+c)}\leq \frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)$
Hoàn toàn tương tự:
$\frac{bc}{a+1}\leq \frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)$
$\frac{ac}{b+1}\leq \frac{ac}{4}\left(\frac{1}{b+c}+\frac{1}{b+a}\right)$
Cộng theo vế các BĐT trên thu được:
$\text{VT}\leq \frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ac}{b+a}+\frac{ac}{b+c}\right)=\frac{1}{4}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{4}(a+b+c)=\frac{1}{4}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
