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