Is it possible to show this expression is positive and bounded between 0 and 1?

[chamilka]

Hi all,
I am working with some beta functions. I want to show that the following is positive and bounded between 0 and 1. Is it possible to show this?

$$ \frac{ \frac{B( a + b , \frac{2}{ c} )}{B(a, \frac{2}{c}) } - \big\{\frac{B( a + b , \frac{1}{ c} )}{B(a, \frac{1}{c}) }\big\}^{2} }{ \frac{B( a + b , \frac{1}{ c} )}{B(a, \frac{1}{c}) } - \big\{\frac{B( a + b , \frac{1}{ c} )}{B(a, \frac{1}{c}) }\big\}^{2} } $$
where a, b and c are positive real numbers.

Thanks for your help.

Here is a image of the above expression.

 

[Sudharaka]

Hi all,
I am working with some beta functions. I want to show that the following is positive and bounded between 0 and 1. Is it possible to show this?

$$ \frac{ \frac{B( a + b , \frac{2}{ c} )}{B(a, \frac{2}{c}) } - \big\{\frac{B( a + b , \frac{1}{ c} )}{B(a, \frac{1}{c}) }\big\}^{2} }{ \frac{B( a + b , \frac{1}{ c} )}{B(a, \frac{1}{c}) } - \big\{\frac{B( a + b , \frac{1}{ c} )}{B(a, \frac{1}{c}) }\big\}^{2} } $$
where a, b and c are positive real numbers.

Thanks for your help.

Here is a image of the above expression.

Hi chamilka, :)

This certainly seem to be bounded between 0 and 1 although I can't think of a way to prove it. Where did you encounter this fraction? You may be interested by the following journal article which includes lots of inequalities for Beta and Gamma functions.

http://rgmia.org/papers/v2n3/begaostr.pdf

Kind Regards,
Sudharaka.

 

[chamilka]

Hi chamilka, :)

This certainly seem to be bounded between 0 and 1 although I can't think of a way to prove it. Where did you encounter this fraction? You may be interested by the following journal article which includes lots of inequalities for Beta and Gamma functions.

http://rgmia.org/papers/v2n3/begaostr.pdf

Kind Regards,
Sudharaka.

Thank you sudharaka. I am checking the reference you suggested. I hope that by considering the Beta integrals it can be proved. Let me to try it! :D

 

[Sudharaka]

Hi chamilka, :)

\[0<x^b<1\mbox{ for }0<x<1\]

\[\therefore x^{a+b}(1-x)^{\frac{1}{c}}<x^{a}(1-x)^{\frac{1}{c}}\mbox{ for }0<x<1\]

\[\Rightarrow \int_0^1 x^{a+b}(1-x)^{\frac{1}{c}}\,dx<\int_0^1 x^{b}(1-x)^{\frac{1}{c}}\,dx\]

\[\Rightarrow \beta\left(a+b,\frac{1}{c}\right)<\beta\left(a, \frac{1}{c}\right)\]

The value of the Beta function is always positive. Therefore,

\[0<\frac{ \beta\left(a+b,\frac{1}{c} \right)}{\beta \left(a, \frac{1}{c}\right)}<1~~~~~~~~~~(1)\]

Using the relationship between Beta and Gamma function it could be shown that,

\[\frac{ \beta\left(a+b,\frac{2}{c} \right)}{\beta \left(a, \frac{2}{c}\right)}\div\frac{ \beta\left(a+b,\frac{1}{c} \right)}{\beta \left(a, \frac{1}{c}\right)}=\frac{\beta\left(a+\frac{2}{c},a+b+\frac{1}{c} \right)}{\beta\left(a+\frac{1}{c},a+b+\frac{2}{c} \right)}\]

Using Theorem 2 on page 7 in the article that I linked in my previous post we can get,

\[\frac{\beta\left(a+\frac{2}{c},a+b+\frac{1}{c} \right)}{\beta\left(a+\frac{1}{c},a+b+\frac{2}{c} \right)}<1\]

\[\therefore \frac{ \beta\left(a+b,\frac{2}{c} \right)}{\beta \left(a, \frac{2}{c}\right)}<\frac{ \beta\left(a+b,\frac{1}{c} \right)}{\beta \left(a, \frac{1}{c}\right)}\]

\[\Rightarrow \frac{ \beta\left(a+b,\frac{2}{c} \right)}{\beta \left(a, \frac{2}{c}\right)}-\left(\frac{ \beta\left(a+b,\frac{1}{c} \right)}{\beta \left(a, \frac{1}{c}\right)}\right)^2<\frac{ \beta\left(a+b,\frac{1}{c} \right)}{\beta \left(a, \frac{1}{c}\right)}-\left(\frac{ \beta\left(a+b,\frac{1}{c} \right)}{\beta \left(a, \frac{1}{c}\right)}\right)^2\]

By (1) it is clear that the right hand side of the inequality is positive. Therefore,

\[\frac{\frac{ \beta\left(a+b,\frac{2}{c} \right)}{\beta \left(a, \frac{2}{c}\right)}-\left(\frac{ \beta\left(a+b,\frac{1}{c} \right)}{\beta \left(a, \frac{1}{c}\right)}\right)^2}{\frac{ \beta\left(a+b,\frac{1}{c} \right)}{\beta \left(a, \frac{1}{c}\right)}-\left(\frac{ \beta\left(a+b,\frac{1}{c} \right)}{\beta \left(a, \frac{1}{c}\right)}\right)^2}<1\]

By (1) the denominator is positive, so it remains to show that the numerator is positive. I'll have to think how to do that...

Kind Regards,
Sudharaka.