Question

Cho các số dương a, b, c, d. Biết \(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}+\dfrac{1}{1+d}\ge3\).

Chứng minh rằng : abcd\(\le\dfrac{1}{81}\)

Answer 1

Từ giả thiết, ta có:

\(\dfrac{1}{1+a}\ge1-\dfrac{1}{1+b}+1-\dfrac{1}{1+c}+1-\dfrac{1}{1+d}=\dfrac{b}{1+b}+\dfrac{c}{c+1}+\dfrac{d}{d+1}\ge3\sqrt[3]{\dfrac{b.c.d}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)

Tương tự:

\(\dfrac{1}{1+b}\ge3\sqrt[3]{\dfrac{cda}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)

\(\dfrac{1}{1+c}\ge3\sqrt[3]{\dfrac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)

\(\dfrac{1}{1+d}\ge3\sqrt[3]{\dfrac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

Nhân vế theo vế 4 BĐT vừa chứng minh rồi rút gọn ta được:

\(abcd\le\dfrac{1}{81}\left(đpcm\right)\)

Answer 2

Mỗi vế trừ đi 4