MT【322】绝对值不等式

已知 $a,b,c\in\mathbb R$，求证：$|a|+|b|+|c|+|a+b+c|\geqslant |a+b|+|b+c|+|c+a|$

分析:不妨设$c=\max\{a,b,c\},\dfrac{a}{c}=x,\dfrac{b}{c}=y$两边同除$|c|$后只需证明

$|x|+|y|+1+|x+y+1|\ge|x+y|+|y+1|+|x+1|$
注意到恒等式$|x|+|y|+|z|=\max\{|x+y+z|,|x+y-z|,|x-y+z|,|x-y-z|\}$,易得.

练习:

设实数$x,y,z$满足
\begin{equation}
\left\{ \begin{aligned}
|x+2y-3z|& \le6\\
|x-2y+3z|&\le6\\
|x-2y-3z|&\le 6\\
|x+2y+3z|&\le6\\
\end{aligned} \right.
\end{equation}
则$|x|+|y|+|z|$的最大值为_____

答案:6

提示:注意到恒等式$|x|+|y|+|z|=\max\{|x+y+z|,|x+y-z|,|x-y+z|,|x-y-z|\}$,易得.