The Inequality

Relevant wiki: Function Graphs

Writing the sigma notation in the series form we get, $\frac {1}{x-1} + \frac {2}{x-2} + \frac {3}{x-3} + \ldots + \frac {70}{x-70} \geq \frac {5}{4}$ .

Now, the most important part of the solution:

Draw the graph of the functions $y = \frac {1}{x-1} + \frac {2}{x-2} + \frac {3}{x-3} + \ldots + \frac {70}{x-70}$ and $y = \frac {5}{4}$ based on logic and deduction.

By logic and deduction, I mean, performing some substitutions in the equation. eg . substituting any $x \in N \leq 70$ , we see that the first function tends towards $\infty$ and so on..

where the set $(1,x_1] \cup (2,x_2] \cup (3,x_3] \ldots (70,x_{70}]$ satisfies the given inequality, then, we have to calculate $x_1-1 + x_2-2 + x_3-3 + \ldots + x_{70}-70$ .

$= x_1 + x_2 + x_3 + \ldots + x_{70} - (1+2+3+\ldots+70)$

Since $x_1, x_2, x_3, \ldots , x_{70}$ are the roots of the equation $\frac {1}{x-1} + \frac {2}{x-2} + \frac {3}{x-3} + \ldots + \frac {70}{x-70} = \frac {5}{4}$ , the value of $x_1 + x_2 + x_3 + \ldots + x_{70}$ can be obtained by applying Vieta's formula for the sum of roots in the above equation.

So, simplifying the above equation, we get, $(x-2)(x-3) \ldots (x-70) + 2 (x-1)(x-3) \ldots (x-70) + \ldots + 70(x-1)(x-2) \ldots (x-69) = \frac{5}{4} (x-1)(x-2)(x-3) \ldots (x-70)$

Transferring everything to the Right hand side, we get the following on the right side,

$\frac{5}{4}x^{70} -\frac{5}{4}x^{69}(1+ 2+ 3+ \ldots + 70) -x^{69}(1+ 2+ 3+ \ldots + 70) + \ldots$ (other terms)

So, sum of roots, $x_1 + x_2 + x_3 + \ldots + x_{70} = \frac{9}{5} (1+ 2+ 3+ \ldots + 70)$

So, the value of $x_1 + x_2 + x_3 + \ldots + x_{70} - (1+2+3+\ldots+70) = \frac{9}{5} (1+ 2+ 3+ \ldots + 70) - (1+ 2+ 3+ \ldots + 70)$

$= \frac{4}{5} (1+ 2+ 3+ \ldots + 70)$

$= \frac{4}{5} \times \frac{71 \times 70}{2}$

$= 4 \times 497$ $= \boxed{1988}$