<s>[INST] Find the minimal value of $\sum\limits_{n=1}^8|x-n|$ 
Find the minimum value of $( |x-1|+|x-2|...+|x-8|) $

My attempt
Using triangle inequality i.e. 
$|x-y|≤|x|+|y|$ $|x-1|=|x-1-0|≤|x-1|+0$ $|x-2|=|x-1-1|≤|x-1|+1$ $|x-3| =|x-1-2| ≤|x-1|+2$
... 
...
...
$|x-8| =|x-1-7| ≤|x-1|+7$
Adding all these inequalities, $( |x-1|+|x-2|...+|x-8|) $$≤$  $8|x-1| + 28$ Clearly the minimum value is 28 at x=1. The answer is wrong. Also it is intuitive that the x must lie somewhere in between 1 to 8 so that the sum of the distances of x from each number is the least.
What am I doing wrong? How else to approach this question. Are there any other ways? 
 [/INST] Use the triangle inequality: $|x+y|\le |x|+|y|$.
$|x-1|+|8-x|\ge |x-1+8-x|=7$,
$|x-2|+|7-x|\ge |x-2+7-x|=5$,
$|x-3|+|6-x|\ge |x-3+6-x|=3$,
$|x-4|+|5-x|\ge |x-4+5-x|=1$,
Summing we get min $16$.
</s>