```Date: Oct 23, 2012 8:54 AM
Author: Jose Carlos Santos
Subject: Re: Ky Fan inequality

On 22-10-2012 2:59, William Elliot wrote:>> Please consider the Ky Fan inequality:>> http://en.wikipedia.org/wiki/Ky_Fan_inequality>>>> I tried to prove it directly (that is, without using some other>> well-known inequality) by induction and already the first step was>> harder than what I expected. This first step is: if 0 < a,b <= 1/2,>> then>>     sqrt(ab)/sqrt((1 - a)(1 - b)) <= (a + b)/(2 - a + b)>> Since all quantities are positive, you're wanting to prove> 	(2 - a + b).sqr ab <= (a + b).sqr (1 - a)(1 - b)> or> 	ab(2 - a + b)^2 <= (a + b)^2 (1 - a)(1 - b).>> Let> 	f(x) = (a + x)^2 (1 - a)(1 - x) - ax(2 - a + x)^2.>> f'(x) = 2(a + x)(1 - a)(1 - x) - (a + x)^2 (1 - a)> 	- (a(2 - a + x)^2 + 2ax)>> f'(0) = 2a(1 - a) - a^2 (1 - a) - a(2 - a)^2> 	= 2a - 2a^2 - a^2 + a^3 - 4a + 4a^2 - a^3> 	= -2a + a^2>> 0 <= f'(0) iff 0 < -2 + a iff 2 < a.> That's not a good sign for b close to 0.>>> What I did was to square both sides and then what I needed to prove>> was that>>>>     (a + b)^2/(2 - a - b)^2 - ab/((1 - a)(1 - b)) >= 0.>> Hey, you changed a sign of b.Yes, but that's because I should have written:    sqrt(ab)/sqrt((1 - a)(1 - b)) <= (a + b)/(2 - (a + b))So, I *still* want to prove that    (a + b)^2/(2 - a - b)^2 - ab/((1 - a)(1 - b)) >= 0.Best regards,Jose Carlos Santos
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