# How do you solve the following question: Show that set of real numbers x, which satisfy the inequality sum_(k=1)^70 k/(x-k) ge 5/4 is a union of disjoint intervals, the sum of whose lengths is 1988?

Sep 13, 2016

See below.

#### Explanation:

$f \left(x\right) = {\sum}_{k = 1}^{70} \frac{k}{x - k}$ is a function with $70$ vertical assymptotes located at $1 , 2 , 3 , \cdots , 70$

$p \left(x\right) = 5 {\Pi}_{k = 1}^{70} \left(x - k\right) - 4 \left({\sum}_{k = 1}^{70} k {\Pi}_{j \ne k}^{70} \left(x - j\right)\right)$ is an associated polynomial to $f \left(x\right)$ with $70$ roots. Due continuity considerations, it's roots are interleaved with the assymptotes so the claimed sum of interval is

$\Delta = {\sum}_{k = 1}^{70} {r}_{k} - {\sum}_{k = 1}^{70} k$ where ${r}_{k}$ are the roots of $p \left(x\right) = 0$

but ${\sum}_{k = 1}^{70} {r}_{k} = {a}_{69}$ the coefficient of ${x}^{69}$ in the $p \left(x\right)$ polynomial. Computing ${a}_{69}$ gives

$5 {a}_{69} = - 5 {\sum}_{k = 1}^{70} k - 4 {\sum}_{k = 1}^{70} k = - 9 {\sum}_{k = 1}^{70} k$

or

${a}_{69} = - \frac{9}{5} \frac{70 \times 71}{2} = - 9 \times 7 \times 71$

Finally, the sum of intervals gives

$\Delta = \left\mid - 63 \times 71 + 35 \times 71 \right\mid = 28 \times 71 = 1988$