prove that the equation has root when x>0

Question

prove that $\left\lvert{x-a_1}\right\rvert$+$\ldots$+$\left\lvert{x-a_i}\right\rvert$=$\frac{n}{2}$has root when x>0

given that ${a_1}$+$\ldots$+${a_i}$=1 , $i=1\ldots n$, and $0\leq a_i\leq1$

I tried finding $$\sum_{i=0}^n{\left\lvert{x-a_i}\right\rvert}$$ I found $$\sum_{i=0}^n{x}=nx$$ and $$\sum_{i=0}^n{a_i}=1$$
so $$\sum_{i=0}^n{\left\lvert{x-a_i}\right\rvert}=|nx-1|$$ I made assisted function $$f(x)=|nx-1|-\frac{n}{2}$$ and I tried to prove that the equation has root when x>0 with the intermediate value theorem
but I don't know how to remove "n" from the equation so i can determent if the final result when assigning x value is negative or positive

Answer 1

Hint:

Define $f(x)=\sum_{j=1}^n|x-a_i|$.

What happens with $f(x)$ for $x \to \infty$? What is the sign of the terms inside the absolute value function? What happens with $x\to 0$? Try other values. If you find a function value below $n/2$ you are done because the function is continuous (see intermediate value theorem).