The solution set of this inequality $\sum_{k=1}^{70}\frac{k}{x-k} \ge \frac{5}{4}$ $$\sum_{k=1}^{70}\frac{k}{x-k} \ge \frac{5}{4}$$
The solution set of this inequality is a union of disjoint intervals and let $S$ be the sum of lengths of intervals of $x$ which satisfy the given inequality, then $S=?$ 
Expanding out gives
$$\frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3} + \dots + \frac{70}{x-70} \ge \frac{5}{4}$$
Also I could just conclude that it will be an equation of degree $70$ but I have no clue on how to approach/solve this problem or what can be the possible values of $x$. Graphing this equation would be really tedious. If anyone can suggest some hints/clues, it would be a great help. Thank You
 A: The function $f(x) = \sum_{k=1}^n \frac{k}{x-k}$ is defined on $\Bbb R \setminus \{ 1, 2, \ldots, n \}$, and

*

*negative on $(-\infty, 1)$,

*strictly decreasing on each interval $(k, k+1)$, $k < n$, with limits $\pm \infty$ at the boundaries of the interval,

*positive and strictly decreasing on $(n, \infty)$ with limits $+\infty$ and zero.

Now fix a number $a > 0$. The set $\{ x \mid f(x) \ge a \}$ is the union of $n$ intervals $(k, x_k]$, where $x_k$ is the unique solution of $f(x) = a$ in $(k, k+1)$ (resp. $(n, \infty)$). The total length is therefore
$$
 S = \sum_{k=1}^n (x_k -k) \, .
$$
Writing the equation $f(x) = a$ in the form
$$
 \prod_{k=1}^n (x-k) - \frac 1a \sum_{k=1}^n \left( k \prod_{l \ne k} (x-l) \right) = 0
$$
and applying Vieta's formulas gives
$$ 
\sum_{k=1}^n x_k = \left( 1 + \frac 1a\right)\sum_{k=1}^n k
$$
and therefore
$$
 S = \frac 1a \sum_{k=1}^n k = \frac{n(n+1)}{2a} \, .
$$
In your case is $n=70$ and $a=5/4$, so that the total length is $S=1988$.

Generalization: The same approach works to compute the solution set of
$$
 \sum_{k=1}^n \frac{b_k}{x-c_k} \ge a
$$
where $a$ and $b_1, \ldots b_n$ are positive real numbers, and $c_1, \ldots, c_n$ are distinct real numbers.
Again the solution set is a union of half-open intervals $[c_k, x_k)$ where $x_1, \ldots, x_n$ are the solutions of
$$
 \prod_{k=1}^n (x-c_k) - \frac 1a \sum_{k=1}^n \left( b_k \prod_{l \ne k} (x-c_k) \right) = 0
$$
and Vieta's formula gives
$$
\sum_{k=1}^n x_k = \sum_{k=1}^n \left( c_k + \frac 1a b_k \right) \, .
$$
The total length of the solution set is therefore
$$
 S = \frac 1a \sum_{k=1}^n b_k \, .
$$
Interestingly, the total length does not depend on the $c_k$.
A: In my former research group, we worked extensively the problem of the $(n-1)$ solutions of the equation
$$f(x)=\sum_{i=1}^n \frac {a_i}{x-b_i}-c=0$$ You could find here our most recent work where are proposed rapid and robust methods of resolution using convex transformations.
Your case is even simpler since $a_i=b_i$ and you are only concerned by the largest root.
The idea is to consider instead the function
$$g(x)=(x-b_n)\sum_{i=1}^n \frac {a_i}{x-b_i}-c(x-b_n)=0$$ and for you case to expand the first term as a series for infinite values of $x$. This gives with your numbers
$$0=-\frac{5
   x}{4}+\frac{5145}{2}-\frac{57155}{x}-\frac{2000425}{x^2}+O\left(\frac{1}{x^3}\right)$$
So, using the first and second term, we have $x_0=2058$ and the first iteration of Newton method applied to this very truncated series will give
$$x_1=\frac{44747735067}{21987445}=2035.1494\cdots$$ while the exact solution is $x=2035.1382\cdots$
