How to prove hypergeometric function inequality

Question

Looking for help on how to prove the following inequality:

$c\,(c-1-x)\,{}_2F_1(1,1\,;\hspace{1pt}c\,;x) \;\,\le\;\, (c-1)\,(c-x)\,{}_2F_1(1,1\,;\hspace{1pt}c+1\,;\,x), \;\;\; 0\le x<1,\; \, c>1 $

I have tried using Gauss contiguous relations, and linear transformation formulae from A&S and G&R. I have looked at the difference of the coefficients in the expansions.

Alternatively:

$$\frac{c\,(c-1-x)}{(c-1)\,(c-x)}\,{}_2F_1(1,1\,;\hspace{1pt}c\,;\,x)\;\;\le\;\; {}_2F_1(1,1\,;\hspace{1pt}c+1\,;x) \;\;\le\;\; {}_2F_1(1,1\,;\hspace{1pt}c\,;\,x),\\\quad\quad\quad\quad \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \;\; 0\le x<1,\; c>1 $$

Also apparently true for $-1<c<1$. This arose in my research trying to prove convexity of the bias of $R^2$ in multiple linear regression.

Thanks.

Answer 1

I figured it out. Using the integral representation, the right side minus the left side of the first inequality above gives:

$\int_0^1 \frac{(1-t)^{c-2}((c+x)t+1)}{1-tx} dt$

which is positive for $0\le x<1$ and $c>-1$.

Answer 2

$$\, _2F_1(1,1;c;x)-\, _2F_1(1,1;c+1;x)=\Gamma (c)\,\sum_{n=1}^\infty\frac{n \, \Gamma (n+1)}{\Gamma (c+n+1)}x^n$$

$$\frac{(c-1)\,(c-x)}{c\,(c-1-x)} \geq 1$$