xi is a infinite sequence of positive reals such that for all n, x1 + x2 + ... + xn ≥ √n. Show that x12 + x22 + ... + xn2 > (1 + 1/2 + 1/3 + ... + 1/n) / 4 for all n.
Solution
Note that this is rather a weak inequality. Taking n = 1, we get x1 >= 1, but (1 + 1/2 + 1/3 + ... + 1/30) < 4, so it is only for n > 30 that we need to consider x2! Of course, weak inequalities can be awkward to prove.
For a + b constant, we minimise a2 + b2 by taking |a - b| as small as possible. So we suspect that the minimum value of x12 + ... + xn2 is when x1 = 1, x2 = √2 - 1, x3 = √3 - √2, x4 = √4 - √3, ... (*). Note that √n - √(n-1) = 1/(√n + √(n-1) > 1/(2√n). So for the values (*) we have x12 + ... + xn2 > (1 + 1/2 + 1/3 + ... + 1/n)/4. So it remains to show that the values (*) do indeed give the minimum.
We use Abel's partial summation formula (whose proof is trivial). Put yn = √n - √(n-1), sn = x1 + x2 + ... + xn, tn = y1 + ... + tn. So we assume sn ≥ tn. Note also that y1 > y2 > y3 > ... . Then the partial summation formula is: ∑ xiyi = ∑ si(yi - yi+1) + snyn+1 (where the sum is taken from 1 to n). We also have ∑ yi2 = ∑ ti(yi - yi+1) + tnyn+1. But each term is not more than the corresponding term in the first equality, so ∑ yi2 ≤ ∑ xiyi.
Now Cauchy-Schwartz gives (∑ xiyi)2 ≤ ∑ xi2 ∑ yi2. Hence ∑ yi2 ≤ ∑ xi2, as required.
© John Scholes
jscholes@kalva.demon.co.uk
23 Aug 2002