The infinite real sequence x1, x2, x3, ... satisfies |xi - xj| ≥ 1/(i + j) for all unequal i, j. Show that if all xi lie in the interval [0, c], then c ≥ 1.
Solution
Suppose that we have 0 ≤ xa(1) < xa(2) < ... < xa(n) ≤ c, for some permutation a(1), a(2), ... , a(n) of 1, 2, ... , n. Then c ≥ xa(n) - xa(1) = (xa(n) - xa(n-1)) + ... + (xa(2) - xa(1)) ≥ 1/(a(n) + a(n-1)) + ... 1/(a(3) + a(2)) + 1/(a(2) + a(1)).
The rest is straightforward. Using Cauchy-Schwartz, we have ( 1/(a(n) + a(n-1)) + ... + 1/(a2) + a(1)) )(a(n) + a(n-1) + a(n-1) + a(n-2) + ... a(3) + a(2) + a(2) + a(1) ) ≥ (1 + 1 + ... + 1)2 = (n - 1)2. Hence c ≥ (n - 1)2/(2(1 + 2 + ... + n)- a(1) - a(n)) = (n - 1)2/(n2 + n - a(1) - a(2) ) ≥ (n - 1)2/(n2 + n - 3) = 1 - (3n - 4)/(n2 + n - 3). That is true for all n, so c ≥ 1.
© John Scholes
jscholes@kalva.demon.co.uk
8 Aug 2003
Last corrected/updated 8 Aug 2003