Russian 2000

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Problem 21

Show that sinn2x + (sinnx - cosnx)2 ≤ 1.

 

Solution

Put s = sin x, c = cos x, so s2 + c2 = 1. Then lhs = 2ns2nc2n + (sn - cn)2 = (2n-2)sncn + s2n + c2n and rhs = 1 = (s2 + c2)n = s2n + c2n + ∑1n-1 nCi s2n-2ic2i. So we have to show that (2n-2)sncn ≤ ∑1n-1 nCi s2n-2ic2i. But that is immediate from AM/GM applied to the 2n-2 terms shck. (Note that there are the same number of terms s2n-2ic2i and s2ic2n-2i and the product of each pair is s2nc2n. Hence the geometric mean is sncn.)

 


 

Russian 2000

© John Scholes
jscholes@kalva.demon.co.uk
4 March 2004
Last corrected/updated 4 Mar 04