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Re: Inequality
Posted:
Aug 1, 2011 1:33 PM


It is easier to prove the stronger statement, where the multiplier is 1 instead of 2^(1r):
a^r  b^r <= a  b^r
Can you prove that one?
Don Coppersmith
> Let a,b >=0 and 0 < r <=1. The inequality (a^r + b^r) > <= 2^(1r) .(a + b)^r holds (the function f(x) = x^r > is concave). How can be proved that: > > a^r  b^r <= 2^(1r) .a  b^r ?



