Áp dụng BĐT Cauchy-Schwarz 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 cho 2 BĐT còn lại :
\(\frac{ca}{b+1}\le\frac{1}{4}\left(\frac{ac}{a+b}+\frac{ac}{b+c}\right);\frac{bc}{a+1}\le\frac{1}{4}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(Q\le\frac{1}{4}\left(\frac{c\left(a+b\right)}{a+b}+\frac{a\left(b+c\right)}{b+c}+\frac{b\left(a+c\right)}{c+a}\right)=\frac{1}{4}\)
\("="\Leftrightarrow a=b=c=\frac{1}{3}\)