Q:

Hypergeometric functions inequality

Let $_2F_1(a,b;c,z)$ be the (Gauss) hypergeometric function, and $m$ and $n$ positive integers.
From a simple plot it looks like
$_2F_1(m+n,1,m+1,\frac{m}{m+n})>\frac{m}{n} \,_2F_1(m+n,1,n+1,\frac{n}{m+n})$
when $m<n$, but how do I prove this?
Hope this is not too trivial, I am not very familiar with such functions.

A:

Using Euler's transformation of the Gauss's hypergeometric function:
$$
    {}_2 F_1\left(a,b; c; z\right) = (1-z)^{c-a-b} \cdot {}_2F_1\left(c-a,c-b; c; z\right)
$$
for $b=1$, $a=m+n$ and $c=m+1$ we have a "closed form" expression:
$$ \begin{eqnarray}
   {}_2 F_1\left(m+n,1; m+1; \frac{m}{m+n}\right) &=& \left(\frac{n}{n+m}\right)^{-n} {}_2F_1\left( 1-n, m; m+1; \frac{m}{m+n} \right) \\ &=& m \left(\frac{n+m}{n}\right)^{n} \cdot \sum_{k=0}^{n-1} \frac{1}{k+m} \frac{(1-n)_k}{k!} \left(\frac{m}{n+m}\right)^k \\
 &=& m \left(\frac{n+m}{n}\right)^{n} \cdot \sum_{k=0}^{n-1} \frac{(-1)^k}{k+m} \binom{n-1}{k} \left(\frac{m}{n+m}\right)^k \\
  &=& m \left(\frac{n+m}{n}\right)^{n} \cdot \left(\frac{n+m}{m}\right)^{m} \int_0^{\frac{m}{n+m}} \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x
\end{eqnarray}
$$
Thus:
$$ 
  \frac{1}{m} {}_2 F_1\left(m+n,1; m+1; \frac{m}{m+n}\right) - 
  \frac{1}{n} {}_2 F_1\left(m+n,1; n+1; \frac{n}{m+n}\right) = 
    \frac{(n+m)^{n+m}}{n^n m^m} \left( \int_0^{\frac{m}{n+m}} \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x - \int_0^{\frac{n}{n+m}} \left(1-x\right)^{m-1} x^{n-1} \mathrm{d} x \right) = 
    \frac{(n+m)^{n+m}}{n^n m^m} \left( \int_0^{\frac{m}{n+m}} \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x - \int_{\frac{m}{n+m}}^1 \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x \right) = \\
\frac{(n+m)^{n+m}}{n^n m^m}  \int_0^1 \operatorname{sgn}\left({\frac{m}{n+m}}-x\right)\left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x
$$
Interpreting $x$ as a beta random variable the last expression becomes:
$$
  \frac{(n+m)^{n+m}}{\Gamma(n+m)} \frac{n^n}{\Gamma(n)} \frac{m^m}{\Gamma(m)} \mathbb{E}\left(  \operatorname{sgn}\left(\mu-X\right) \right)  = 
   \frac{(n+m)^{n+m}}{\Gamma(n+m)} \frac{n^n}{\Gamma(n)} \frac{m^m}{\Gamma(m)} \left( \mathbb{P}\left( X < \mu \right) - \mathbb{P}\left( X > \mu \right)\right)
$$
where $\mu = \mathbb{E}(X) = \frac{m}{m+n}$, where $X$ follows $\mathrm{Be}(m,n)$. Given that $m<n$, random variable $X$ is positively skewed, and the expression is positive.

