\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
\(\Rightarrow\dfrac{1}{a}=\left(\dfrac{1}{2}-\dfrac{1}{b}\right)+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{b-2}{2b}+\dfrac{c-2}{2c}\)
Dễ dàng chứng minh \(\dfrac{b-2}{2b},\dfrac{c-2}{2c}\) là các số dương.
Áp dụng BĐT Cauchy cho 2 số dương ta có:
\(\dfrac{b-2}{2b}+\dfrac{c-2}{2c}\ge2\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{4bc}}=\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{bc}}\)
\(\Rightarrow\dfrac{1}{a}\ge\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{bc}}\left(1\right)\)
CMTT ta có: \(\left\{{}\begin{matrix}\dfrac{1}{b}\ge\sqrt{\dfrac{\left(c-2\right)\left(a-2\right)}{ca}}\left(2\right)\\\dfrac{1}{c}\ge\sqrt{\dfrac{\left(a-2\right)\left(b-2\right)}{ab}}\left(3\right)\end{matrix}\right.\)
\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow\dfrac{1}{abc}\ge\dfrac{\left(a-2\right)\left(b-2\right)\left(c-2\right)}{abc}\)
\(\Rightarrow\left(a-2\right)\left(b-2\right)\left(c-2\right)\le1\left(đpcm\right)\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=b=c\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\end{matrix}\right.\Leftrightarrow a=b=c=3\)