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? 

Your answer is wrong, but there is nothing contradictory with your derivation. You used the inequality $(|x-1|+\cdots+|x-8|)\leq 8|x-1|+28$, which is loose at the optimal solution of $x$. In other words, $x=1$ minimizes $8|x-1|+28$ but not $(|x-1|+\cdots+|x-8|)$, because of the inequality.
A principled way to approach this question is by case analysis on $x$: analyze $x\leq 0$, $0\leq x\leq 1$, $1\leq x\leq 2$, $\cdots$, $x\geq 8$. The objective function would become linear in each of the above-mentioned regions, and the minimizer of a linear function over an interval $[a,b]$ is always attained on the boundary of the interval. Therefore, you only need to evaluate your objective at $x=1,\cdots,x=8$, and one of them would be the minimizer. It is clear that both $x=4$ and $x=5$ minimize your function, with a optimal value of 16.
