\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=1\Rightarrow1-\dfrac{1}{1+a}=\dfrac{1}{1+b}+\dfrac{1}{1+c}\)
\(\Rightarrow\dfrac{a}{1+a}\ge\dfrac{1}{1+b}+\dfrac{1}{1+c}\ge2\sqrt{\dfrac{1}{\left(1+b\right)\left(1+c\right)}}\) (1)
Tương tự ta có:
\(\dfrac{b}{1+b}\ge2\sqrt{\dfrac{1}{\left(1+a\right)\left(1+c\right)}}\) (2)
\(\dfrac{c}{1+c}\ge2\sqrt{\dfrac{1}{\left(1+a\right)\left(1+b\right)}}\) (3)
Nhân vế (1);(2);(3):
\(\Rightarrow\dfrac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
\(\Rightarrow abc\ge8\)
Dấu "=" xảy ra khi \(a=b=c=2\)