Áp dụng BĐT \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) ta có:
\(\frac{ab}{c+1}=\frac{ab}{\left(a+c\right)\left(b+c\right)}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
Tương tự ta có:
\(\frac{bc}{a+1}\le\frac{1}{4}\left(\frac{bc}{b+a}+\frac{bc}{c+a}\right);\frac{ac}{b+1}\le\frac{1}{4}\left(\frac{ac}{a+b}+\frac{ac}{c+b}\right)\)
Cộng theo vế ta được:
\(\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ac}{b+1}\le\frac{1}{4}\left[\left(\frac{ab}{b+c}+\frac{ac}{c+b}\right)+\left(\frac{ab}{a+c}+\frac{bc}{c+a}\right)+\left(\frac{bc}{b+a}+\frac{ac}{a+b}\right)\right]\)
\(=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)
Dấu "=" khi \(a=b=c=\frac{1}{3}\)
