α is a non-negative real. x is a real satisfying (x + 1)2 ≥ α(α + 1). Is x2 ≥ α(α - 1)?
Solution
Yes.
Easy.
Suppose α(α+1) = (x+1)2 and x >= 0. Then since (x + 1/2)(x + 3/2) = (x + 1)2 - 1/4, we have α > x + 1/2. Hence α(α - 1) = α(α + 1) - 2α < (x + 1)2 - 2(x + 1/2) = x2.
If α(α+1) < (x+1)2 and x ≥ 0, then take β > α with β(β + 1) = (x+1)2. Clearly α - 1 < β - 1 and since α is positive, α(α - 1) < α(β - 1) < β(β - 1) which is < x2.
So it just remains to consider the case x < 0. But in this case -|x| - 1 < x + 1 < |x| + 1, so (x + 1)2 < (|x| + 1)2 and x2 = |x|2, and the result follows from the result for α, |x|.
© John Scholes
jscholes@kalva.demon.co.uk
12 Dec 1998