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Topic: Ky Fan inequality
Replies: 5   Last Post: Oct 25, 2012 5:02 AM

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William Elliot

Posts: 1,609
Registered: 1/8/12
Re: Ky Fan inequality
Posted: Oct 21, 2012 9:59 PM
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On Sun, 21 Oct 2012, José Carlos Santos 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.

> So, I turned the LHS of this expression into a rational expression,
> whose denominator is clearly >= 0 and whose numerator is
>
> a^2 - a^3 + b^2 - b^3 + a^2b + ab^2 - 2ab. (*)
>
> Since the original inequality is actually an equality when a = b, this
> is also true for (*) and so I divided (*) by a - b, getting
>
> a - a^2 - b + b^2 = a - b - (a^2 - b^2) = (a - b)(1 - a - b).
>
> So, (*) = (a - b)^2(1 - a - b), which is clearly >= 0.
>
> I have the feeling that what I did was too complicated. Can anyone see
> a shorter way of proving this (again, avoiding the use of well-known
> inequalities)?
>
> Best regards,
>
> Jose Carlos Santos
>




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