\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2\)
\(\Leftrightarrow\frac{1}{1+a}=1-\frac{1}{1+b}+1-\frac{1}{1+c}\)
\(\Leftrightarrow\frac{1}{1+a}=\frac{b}{1+b}+\frac{c}{1+c}\ge2\sqrt{\frac{bc}{\left(1+b\right)\left(1+c\right)}}\) (BĐT Cosi)
Tương tự ta có \(\hept{\begin{cases}\frac{1}{1+b}\ge2\sqrt{\frac{ac}{\left(1+a\right)\left(1+c\right)}}\\\frac{1}{1+c}\ge2\sqrt{\frac{ab}{\left(1+a\right)\left(1+b\right)}}\end{cases}}\)
Nhân vế theo vế \(\Rightarrow\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge8\sqrt{\frac{a^2b^2c^2}{\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}}=\frac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
\(\Rightarrow abc\le\frac{1}{8}\)
Dấu "=" xảy ra khi a=b=c=\(\frac{1}{2}\)
Nguồn:Hoàng Phương
