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Re: Ky Fan inequality
Posted:
Oct 23, 2012 8:54 AM


On 22102012 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 >> wellknown 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



