\(a,b,c>0\)
\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=2\)
\(\Leftrightarrow\dfrac{1}{1+a}=\left(1-\dfrac{1}{1+b}\right)+\left(1-\dfrac{1}{1+c}\right)\)
\(\Leftrightarrow\dfrac{1}{1+a}=\dfrac{b}{1+b}+\dfrac{c}{1+c}\)
Theo BĐT AM-GM (Caushy) ta có:
\(\dfrac{1}{1+a}=\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge2\sqrt{\dfrac{b}{1+b}.\dfrac{c}{1+c}}>0\left(1\right)\)
Tương tự:
\(\dfrac{1}{1+b}\ge2\sqrt{\dfrac{c}{1+c}.\dfrac{a}{1+a}}>0\left(2\right)\)
\(\dfrac{1}{1+c}\ge2\sqrt{\dfrac{a}{1+a}.\dfrac{b}{1+b}}>0\left(3\right)\)
Nhân theo vế (1), (2), (3) ta có:
\(\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge8.\dfrac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}>0\)
\(\Leftrightarrow8abc\le1\)
\(\Leftrightarrow abc\Leftrightarrow\dfrac{1}{8}\left(đpcm\right)\)
Đẳng thức xảy ra \(\Leftrightarrow\dfrac{1}{1+a}=\dfrac{1}{1+b}=\dfrac{1}{1+c}=\dfrac{2}{3}\Leftrightarrow a=b=c=\dfrac{1}{2}\)
