Show that x1/(1 + x12) + x2/(1 + x12 + x22) + ... + xn/(1 + x22 + ... + xn2) < √n for all real numbers xi.
Solution
Using Cauchy ∑ aibi ≤ √(∑ ai2) √(&sum bi2) with ai = 1, bi = xi/(1+x12+x22+...+xi2) we have lhs ≤ rhs √(∑ bi2).
Now use b12 ≤ x12/(1+x12) = 1 - 1/(1+x12)
© John Scholes
jscholes@kalva.demon.co.uk
20 Oct 2003
Last corrected/updated 20 Oct 03