Không mất tính tổng quát ta giả sử:
\(0\le a\le b\le c\le4\)
Ta có: \(\dfrac{1}{\left(a-b\right)^2}+\left(b-a\right)+\left(b-a\right)\ge3\)(1)
\(\dfrac{1}{\left(b-c\right)^2}+\left(c-b\right)+\left(c-b\right)\ge3\left(2\right)\)
\(\dfrac{1}{\left(c-a\right)^2}+\dfrac{\left(c-a\right)}{8}+\dfrac{\left(c-a\right)}{8}\ge\dfrac{3}{4}\left(3\right)\)
Cộng (1), (2), (3) vế theo vế rồi rút gọn ta được.
\(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(c-a\right)^2}+\dfrac{9c-9a}{4}\ge\dfrac{27}{4}\)
\(\Leftrightarrow\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(c-a\right)^2}\ge\dfrac{27}{4}+\dfrac{9a-9c}{4}\)
\(\ge\dfrac{27}{4}+\dfrac{9.0-2.9}{4}=\dfrac{9}{4}\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}a=0\\b=1\\c=2\end{matrix}\right.\)
