Is this function negative?

[george...@gmail.com]

Hello,

I am trying to show that the following function is always negative,
for any m and n >0:

f(m,n) := trigamma(n)*trigamma(n+m) + trigamma(m)*trigamma(n+m) -
trigamma(m)*trigamma(n)

where the trigamma function is monotone decreasing on R+ (http://
mathworld.wolfram.com/TrigammaFunction.html).

I can see numerically that it is so but can't figure out how to
demonstrate it.

Any hint would be very welcome!

Thank you very much in advance,
George.

[Graham Jones]

<george...@gmail.com> wrote in message
news:5486cba1-d6f4-47bc...@k30g2000hse.googlegroups.com...

This is just a hint, and it may be useless... Your inequality follows from

1/trigamma(n) + 1/trigamma(m) < 1/trigamma(n+m)

and g(x) = 1/trigamma(x) looks convex.

Graham

[george...@gmail.com]

On 23 juil, 20:25, "Graham Jones" <x...@x.x> wrote:
> This is just a hint, and it may be useless... Your inequality follows from
>
> 1/trigamma(n) + 1/trigamma(m) < 1/trigamma(n+m)

This is very insightful and much more elegant than what I had tried!

> and g(x) = 1/trigamma(x) looks convex.

I am not still not quite sure how I can go formally from convexity to
f(m,n)>f(m)+f(n) but I'll look into that.

Thank you very much for your help.

George.

[Graham Jones]

<george...@gmail.com> wrote in message
news:21a10126-04d7-4391...@x29g2000prd.googlegroups.com...

You also need g(0) = 0, but that is easy. Then from the definition of
(strict) convexity

g((n/r)0 + (m/r)r) < (n/r)g(0) + (m/r)g(r)

where r = m+n, which simplifies to

g(m) < (m/r)g(r).

Likewise

g(n) < (n/r)g(r).

Now add these two up. Showing g(x) = 1/trigamma(x) is convex is the hard
part.

Graham