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

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Jose Carlos Santos

Posts: 4,873
Registered: 12/4/04
Re: Ky Fan inequality
Posted: Oct 23, 2012 8:58 AM
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On 23-10-2012 1:33, Ken Pledger wrote:

>> .... This first step is: if 0 < a,b <= 1/2,
>> then
>>
>> sqrt(ab)/sqrt((1 - a)(1 - b)) <= (a + b)/(2 - a + b)
>>
>> .... Can anyone see
>> a shorter way of proving this (again, avoiding the use of well-known
>> inequalities)? ....

>
>
> I assume your (2 - a + b) is a typo for (2 - a - b).


Sure. Thanks.

> Transform the problem a bit. You want to prove that
>
> (a + b)/sqrt(ab) >= ((1 - a) + (1 - b))/sqrt((1 - a)(1 - b))
>
> i.e. sqrt(a/b) + sqrt(b/a) >=
> sqrt((1 - a)/(1 - b)) + sqrt((1 - b)/(1 - a))
>
> i.e. sqrt(a/b) - sqrt((1 - a)/(1 - b)) >=
> sqrt((1 - b)/(1 - a)) - sqrt(b/a).
>
> If you put each side over the appropriate common denominator, then the
> numerators come out the same on both sides. Assuming wolog a >= b,
> it's easy to show that the common numerator is positive, so cancel it.
> Then the necessary inequality between the denominators is also
> elementary.


Cute. Thanks.

Best regards,

Jose Carlos Santos




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