Áp dụng bđt |a|+|b|+|c|+|d| \(\ge\)|a+b+c+d| ta có:
B = |x-2016|+|x-2015|+|x-2014|+|x-2013|+|x-2012|+2016
B = |2016-x|+|2015-x|+|x-2014|+|x-2013|+|x-2012|+2016 \(\ge\) |(2016-x)+(2015-x)+0+(x-2013)+(x-2012)|+2016 = |6|+2016 = 6+2016 = 2022
Dấu "=" xảy ra khi \(\left\{\begin{matrix}x-2015\le0\\x-2014=0\\x-2013\ge0\end{matrix}\right.\) => x = 2014
Ta có: \(\left|x-2016\right|\ge0\forall x\in R\)
\(\left|x-2015\right|\)\(\ge0\forall x\in R\)
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=> |x-2016|+|x-2015|+|x-2014|+|x-2013|+|x-2012| \(\ge0\forall x\in R\)
=> |x-2016|+|x-2015|+|x-2014|+|x-2013|+|x-2012| + 2016 \(\ge0\forall x\in R\)
Dấu "=" xảy ra khi \(\left|x-2016\right|=0\); .....; \(\left|x-2012\right|=0\) Với \(\left|x-2016\right|=0\) => x = \(2016\) Với \(\left|x-2015\right|=0\) => x = 2015 Với \(\left|x-2014\right|=0\) => x = 2014 Với \(\left|x-2013\right|=0\) => x = 2013 Với \(\left|x-2012\right|=0\) => x = 2012 Vậy GTNN của B = 2016 khi x \(\in\) \(\left\{2016;2015;2014;2013;2012\right\}\)
