Cho mình sửa lại đề:
CMR:(x-2)(y-2)(z-2)\(\le\)1
Đặt a=x-2, b=y-2, c=x-2,
Theo đề bài, ta có:
\(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)
<=>\(\dfrac{1}{a+2}=1-\dfrac{1}{b+2}-\dfrac{1}{c+2}\)
<=>\(\dfrac{1}{a+2}=\left(\dfrac{1}{2}-\dfrac{1}{b+2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{c+2}\right)\)
<=>\(\dfrac{1}{a+2}=\dfrac{b}{2\left(b+2\right)}+\dfrac{c}{2\left(c+2\right)}\)
Ta có:
\(\dfrac{b}{2\left(b+2\right)}+\dfrac{c}{2\left(c+2\right)}\ge2\sqrt{\dfrac{bc}{4\left(b+2\right)\left(c+2\right)}}=\sqrt{\dfrac{bc}{\left(b+2\right)\left(c+2\right)}}\)=>\(\dfrac{1}{a+2}\ge\sqrt{\dfrac{bc}{\left(b+2\right)\left(c+2\right)}}\)(1)
Tương tự, ta cũng sẽ có:
\(\dfrac{1}{b+2}\ge\sqrt{\dfrac{ac}{\left(a+2\right)\left(c+2\right)}}\)(2)
\(\dfrac{1}{c+2}\ge\sqrt{\dfrac{ab}{\left(a+2\right)\left(b+2\right)}}\)(3)
Lấy (1) , (2), (3) nhân lại với nhau,ta sẽ có:
\(\dfrac{1}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\ge\sqrt{\dfrac{a^2b^2c^2}{\left(a+2\right)^2\left(b+2\right)^2\left(c+2\right)^2}}\)
=>\(\dfrac{1}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\ge\dfrac{abc}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)
=>\(1\ge abc\) hay \(abc\le1\)
=>(x-2)(y-2)(z-2)\(\le1\)
