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