Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B=\left|x-2011\right|+\left|x-2\right|\)
\(=\left|x-2011\right|+\left|2-x\right|\)
\(\ge\left|x-2011+2-x\right|=2009\)
Xảy ra khi \(2\le x\le2011\)
\(\left|x-2011\right|+\left|x-2\right|=\left|x-2011\right|+\left|2-x\right|\)
Áp dụng bđt:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow\left|x-2011\right|+\left|2-x\right|\ge\left|x-2011+2-x\right|\)
\(\Rightarrow\left|x-2011\right|+\left|2-x\right|\ge2009\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2011\ge0\Rightarrow x\ge2011\\2-x\ge0\Rightarrow x\le2\end{matrix}\right.\\\left\{{}\begin{matrix}x-2011< 0\Rightarrow x< 2011\\2-x< 0\Rightarrow x>2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow2< x< 2011\)
