Show that set of real numbers $x$ which satisfy the inequality
$\sum_{k=1}^{70}\frac{k}{x-k}\geq\frac{5}{4}$
is a union of disjoint, half-open intervals, the sum of whose lengths is 1988.
My solution:
Let the function on the LHS of the inequality be $f(x)$, and consider the roots of the equation $f(x)=\frac{5}{4}$
Since $f(x)<0$ for $x<1$, any roots of this equation must occur with $x\geq 1$.
Now $f(x)$ has 70 discontinuities, namely at $x=1,2..,70$. Also, its first derivative is negative for all points at which it is defined, and it is clear that its range is infinite in both directions between any two of its discontinuities. Therefore, by IVT, it must have a root between each pair of discontinuities.
Furthermore, since $f(x)\to0$ as $x\to\infty$, it must have a single root at some value of $x>70$ (only one because it is strictly decreasing).
We see from this that the required subset is indeed a disjoint union of half-open intervals.
Now multiplying through the equation by
$\prod_{k=1}^{70} (x-k)$ and rearranging leads to a degree 70 polynomial in $x$ with leading coefficient $\frac{5}{4}$, and sum of roots (by Vieta's formula) being the 70th triangular number with a factor of $\frac{\frac{5}{4}+1}{\frac{5}{4}}=\frac{9}{5}$
which gives 4473
Now to get the sum of the required lengths from here, we simply subtract the 70th triangular number, i.e. the overcounted intervals before each discontinuity.
Thus, 4473 - 2485 = 1988.
QED
I'm wondering if this would get full credit, and welcome any suggestions for improvement of substance or style. Thanks.