Find solution set of $200x=\sum_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right| \right)}$ Let $f(x)=\sum\nolimits_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right| \right)}$. How can I find all the roots when $f(x)=200x$?
I first tried to use the Triangle Inequality as $|a|+|b|>|a+b|$ but this doesn't make sense. I then used a graphing device and found the solution set is $\{x:x\geq 100\}$.
My question: is there any shortcut to attack such a problem ?
 A: Clearly, $x\geq 0.$ Rewrite the equation $\;200x = \sum_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right| \right)}$ as
$$\sum_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right|-2x \right)}=0$$
Consider two possible cases:

*

*If $\;x\geq k\;$ for some $k,\;$ then $\;| x-k |+| x+k |-2x=0.$

*If $\;0 \leq x < k\;$ for some $k,\;$ then $\;| x-k |+| x+k |-2x=2k-2x>0.$
From 1. and 2. it follows that
$$\sum_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right|-2x \right)}=0 \iff \big(x\geq k \;\;\text{for all} \;\; k\big),$$ that is, if $x\geq 100.$
A: If you are looking for
$$200x = \sum_{k=1}^{100}{\left( \left| x-k \right|+\left| x+k \right| \right)}$$
then you can notice that  $|x+k| \geq 0$ and $|x-k| \geq$ $0$ for all $k$ when $x \geq 100$ so you have your equality.
When $x \leq 0$ instead you can't have equality because lhs is negative or zero while rhs is positive. When $0 < x < 1$, instead you have the rhs greater than the lhs.
Remains to investigate $1 < x< 100$,  say you have $q \leq x  < q+1$ with $q \in \{1,2,...,99\}$, then you can rewrite the rhs as
$$\sum_{k=1}^{q}{\left( \left| x-k \right|+\left| x+k \right| \right)} + \sum_{k=q+1}^{100}{\left( \left| x-k \right|+\left| x+k \right| \right)}  =  2qx +100\cdot101-q(q+1)$$
but
$$200x <  2qx +100\cdot101-q(q+1) \iff x  < \frac{q+101}{2}$$
and this is always the case when  $q \leq x  < q+1$.
So it's proven that  the solution set is $\{x:x\geq 100\}$.
