xi and yi are positive reals with ∑1n xi = ∑1n yi. Show that ∑1n xi2/(xi + yi) ≥ (∑1n xi)/2.
Solution
We use Cauchy-Schwartz: ∑ (x/√(x+y) )2 ∑ (√(x+y) )2 ≥ (∑ x )2. So ∑ x2/(x+y) >= (∑ x)2/(∑(x+y) = 1/2 ∑ x.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002