Show that the set of real numbers x which satisfy the inequality:

1/(x - 1) + 2/(x - 2) + 3/(x - 3) + ... + 70/(x - 70) ≥ 5/4

is a union of disjoint intervals, the sum of whose lengths is 1988.

**Solution**

Let f(x) = 1/(x - 1) + 2/(x - 2) + 3/(x - 3) + ... + 70/(x - 70). For any integer n, n/(x - n) is strictly monotonically decreasing except at x = n, where it is discontinuous. Hence f(x) is strictly monotonically decreasing except at x = 1, 2, ... , 70. For n = any of 1, 2, ... , 70, n/(x - n) tends to plus infinity as x tends to n from above, whilst the other terms m/(x - m) remain bounded. Hence f(x) tends to plus infinity as x tends to n from above. Similarly, f(x) tends to minus infinity as x tends to n from below. Thus in each of the intervals (n, n+1) for n = 1, ... , 69, f(x) decreases monotonically from plus infinity to minus infinity and hence f(x) = 5/4 has a single foot x_{n}. Also f(x) ≥ 5/4 for x in (n, x_{n}] and f(x) < 5/4 for x in (x_{n}, n+1). If x < 0, then every term is negative and hence f(x) < 0 < 5/4. Finally, as x tends to infinity, every term tends to zero, so f(x) tends to zero. Hence f(x) decreases monotonically from plus infinity to zero over the range [70, infinity]. Hence f(x) = 5/4 has a single root x_{70} in this range and f(x) >= 5/4 for x in (70, x_{70}] and f(x) < 5/4 for x > x_{70}. Thus we have established that f(x) ≥ 5/4 for x in any of the disjoint intervals (1, x_{1}], (2, x_{2}], ... , (70, x_{70}] and f(x) < 5/4 elsewhere.

The total length of these intervals is (x_{1} - 1) + ... + (x_{70} - 70) = (x_{1} + ... + x_{70}) - (1 + ... + 70). The x_{i} are the roots of the 70th order polynomial obtained from 1/(x - 1) + 2/(x - 2) + 3/(x - 3) + ... + 70/(x - 70) = 5/4 by multiplying both sides by (x - 1) ... (x - 70). The sum of the roots is minus the coefficient of x^{69} divided by the coefficient of x^{70}. The coefficient of x^{70} is simply k, and the coefficient of x^{69} is - (1 + 2 + ... + 70)k - (1 + ... + 70). Hence the sum of the roots is (1 + ... + 70)(1 + k)/k and the total length of the intervals is (1 + ... + 70)/k = 1/2 70·71 4/5 = 28·71 = 1988.

Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

© John Scholes

jscholes@kalva.demon.co.uk

25 Oct 1998