Is the Gaussian hypergeometric function ${}_2F_1$ bigger than 1?

Question

I want to know if the Gaussian hypergeometric function ${}_2F_1$ satisfies

$$ {}_2F_1(a,b;c;x) \geq 1 $$ for all $a,b\leq 0$, $c > 0$, $0 < x < 1$. (If not, how would we restrict $a,b$ further for the inequality to hold.)

I think it is true based on plots with Mathematica but I have not managed to prove it. Any ideas ?

Note: this inequality holds if $a,b \geq 0$, but the case $a,b\leq 0$ seems less clear.

Answer 1

It is known the hypergeometric function is the solution to following ODE

$$(1-x)x w'' + (c - (a + b + 1)x) w' - ab w = 0$$

with $w(0)=1$ and $w'(0)=\frac{ab}{c}$. If $a$ or $b$ is zero, this equation can be solved explicitly. Let's consider the generic case $a < 0, b < 0$. Now $w'(0) > 0$ with OP's condition.

Suppose $z_0\in (0, 1)$ is the first zero (if exists) of $w'$, then $w$ is increasing over $(0, z_0)$.

$$w''(z_0) z_0(1 - z_0) = ab w(z_0) > 0$$

which means $w$ has a local minimum at $z_0$, which means $w'(z) > 0$ for $z=z_0^{+}$. Then you can use the same idea to look for the next zero (if exists) on $(z_0, 1)$.

Note there are only a finite number of zeros on $(0, 1)$ for $w'$.

Answer 2

Letac, G. and Piccioni, M. (2012) 'Random continued fractions with beta hypergeometric distribution.' Ann. Probab. 40, number 3, 1105-1134.

In that paper Theorem 2.1 gives the necessary and sufficient answer to a close question, namely to positiveness of $_2 F_1$ on $(0,1).$ It is due to Felix Klein. Our paper contains the reference to Klein.