☘ Áp dụng bất đẳng thức AM - GM
\(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}+\dfrac{d}{1+d}=1\)
\(\Leftrightarrow1-\dfrac{a}{1+a}=\dfrac{b}{1+b}+\dfrac{c}{1+c}+\dfrac{d}{1+d}\)
\(\Rightarrow\dfrac{1}{1+a}\ge3\sqrt[3]{\dfrac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)
☘ Tương tự, ta cũng có:
\(\dfrac{1}{1+b}\ge3\sqrt[3]{\dfrac{acd}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)
\(\dfrac{1}{1+c}\ge3\sqrt[3]{\dfrac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\dfrac{1}{1+d}\ge3\sqrt[3]{\dfrac{abc}{\left(1+a\right)\left(1+c\right)\left(1+b\right)}}\)
☘ Nhân vế theo vế
\(\Rightarrow\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge\dfrac{81abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\)
\(\Rightarrow abcd\le\dfrac{1}{81}\)
☘ Dấu "=" xảy ra khi \(a=c=b=d=\dfrac{1}{3}\)
⚠ Nguồn: https://hoc24.vn/hoi-dap/question/463672.html