Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

 Messages: [ Previous | Next ]
 Ken.Pledger@vuw.ac.nz Posts: 1,412 Registered: 12/3/04
Re: Ky Fan inequality
Posted: Oct 22, 2012 8:33 PM

In article <aei7kuF620vU1@mid.individual.net>,
José Carlos Santos <jcsantos@fc.up.pt> 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).

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.

Ken Pledger.

Date Subject Author
10/21/12 Jose Carlos Santos
10/21/12 William Elliot
10/23/12 Jose Carlos Santos
10/25/12 William Elliot
10/22/12 Ken.Pledger@vuw.ac.nz
10/23/12 Jose Carlos Santos