Ta có:
\(\frac{1}{1+a}=2-\frac{1}{1+b}-\frac{1}{1+c}=\left(1-\frac{1}{1+b}\right)+\left(1-\frac{1}{1+c}\right)\ge\frac{b}{1+b}+\frac{c}{1+c}\ge2\sqrt{\frac{bc}{\left(1+b\right)\left(1+c\right)}}\)
Tương tự:
\(\frac{1}{1+b}\ge2\sqrt{\frac{ac}{\left(1+a\right)\left(1+c\right)}}\)
\(\frac{1}{1+c}\ge2\sqrt{\frac{ab}{\left(1+a\right)\left(1+b\right)}}\)
=> \(\frac{1}{1+a}.\frac{1}{1+b}.\frac{1}{1+c}\ge\frac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
=> \(abc\le\frac{1}{8}\)
"=" xảy ra <=> a = b = c = 1/2
Vậy max P = abc = 1/8 đạt tại a = b = c =1/2
