I am working with a certain Gauss hypergeometric function, $\_{2}F\_{1}(a,a-b;2a;1-z)$, where $a, b \in {\mathbb R}$ with $a, a-b > 0$ and $0<b<1$.


It appears that $\left| \_{2}F\_{1}(a,a-b;2a;1-z) \right| \geq 1$ for all $|z| \leq 1$.


More generally, it seems that
$$
\left| \_{2}F\_{1}(a,b;c;1-z) \right| \geq 1
$$
for all $|z| \leq 1$ when $a,b,c \in {\mathbb R}$ with $a,b>0$ and $c \geq \max( a,b)$.


For example, if $0 \leq z \leq 1$, then this is true since all the coefficients in the expansion of $\_{2}F\_{1}(a,b;c;z)$ are positive.


But I have been unable to prove the more general result for all $z$ with $|z| \leq 1$ myself or find a proof in the literature.


Would anyone have any ideas on this, or references I may have missed?



lower bound for absolute value of a hypergeometric function
As I said, the problem is to show that the absolute value of 
$$
z\mapsto\int\_0^\infty t^\alpha(t+1)^{-\beta}(t+z)^{-\gamma}\frac {dt}t
$$
attains its minimum in $\Omega=\{z:|z|\le 1, z\notin[-1,0]\}$ at $z=1$. Note that we are free to change the contour of integration to any curve that stays in the open angle bounded by the rays $\{-\tau w:\tau>0\}$ and $\{-\tau:\tau>0\}$ containing the positive semi-axis. So, I'll immediately change it to the ray $\{\tau\sqrt w:\tau>0\}$, which reduces the problem to the following. Let $z,w\in\mathbb C, \Re z,\Re w>0, zw\in\mathbb R\_+$. Then
$$
\left|\int\_0^\infty t^\alpha(t+w)^{-\beta}(t+z)^{-\gamma}\frac {dt}t\right|\ge
\left|\int\_0^\infty t^\alpha(t+|w|)^{-\beta}(t+|z|)^{-\gamma}\frac {dt}t\right|\,.
$$
Note that this inequality is immediate if $\beta=\gamma$ and $z=\bar w$ (which corresponds to $c=2a$, $|z|=1$ in the original formulation) because the integrand on the left is then positive and obviously greater than the one on the right. However, in general the integrand on the left oscillates so we cannot take advantage of its being large in absolute value immediately. 


Still it suggests the following idea: change the contour (we are now free to choose any curve $\Gamma$ connecting $0$ and $\infty$ in the right half-plane) so that the argument of the product $F(t)=t^\alpha(t+w)^{-\beta}(t+z)^{-\gamma}$ stays fixed (say, $\theta$) on $\Gamma$. If, in addition, we knew that $\Gamma$ crossed each circle centered at the origin just once, we would be able to write
$$
\left|\int\_\Gamma f(t)\frac{dt}t\right|\ge \Re\left[\int\_\Gamma e^{-i\theta}F(t)\frac{dt}t\right]=\int\_\Gamma |F(t)|d(\log|t|)
\\
\ge\int\_\Gamma |t|^\alpha(|t|+|w|)^{-\beta}(|t|+|z|)^{-\gamma}d(\log|t|)
=\int\_0^\infty t^\alpha(t+|w|)^{-\beta}(t+|z|)^{-\gamma}\frac{dt}{t}\,,
$$
making everything obvious again. However we face two problems here: we have to demonstrate the existence of such a curve and the circle crossing property. It turns out that the fixed argument condition determines the curve uniquely and we'll define it shortly. However, as far as proving the circle crossing property is concerned (numerical experiments indicate it should be true), I have lost my competition with the tree stump in my backyard: we both couldn't do it after two days of thinking but the stump was obviously much more concentrated on the problem since it wasn't getting distracted by anything (hey, it was so deep in thought that it even didn't move for 2 days!), so we'll have to circumvent it.


The next thing we need is that for $\Re z>0$, the function 
$$
[-\pi/2,\pi/2]\ni \theta\mapsto |re^{i\theta}+z|^2=r^2+|z|^2+2sr\Im z+2\sqrt{1-s^2}r\Re z
$$ 
is a non-negative strongly (in the sense of non-degenerate second derivative) concave function of $s=\sin\theta\in[-1,1]$. It follows that the function 
$$
\theta\mapsto -\beta\log|w+re^{i\theta}|-\gamma\log|z+re^{i\theta}|
$$ 
is a strongly convex function of $s$. Moreover, when going slightly away from the endpoints $\theta=\pm\pi/2$ it obviously decreases (both absolute value increase). Thus on each semicircle $S\_r=\{re^{i\theta}:\theta\in[-\frac\pi 2,\frac\pi 2]\}\ni t$ the function $L(t)=\alpha\log|t|-\beta\log|w+t|-\gamma\log|z+t|$ has a unique strong minimum $t(r)$, which, thereby, depends on $r$ continuously, and when you move from $t(r)$ towards either endpoint of $S\_r$, it increases monotonically. Clearly, $L(t(r))\to -\infty$ for $r\to 0$ and $r\to\infty$ (that's where we use the condition $\alpha<\beta+\gamma$), so there is at least one maximum $r\_\*$ of $r\mapsto L(t(r))$. Any such maximum corresponds to a non-degenerate saddle point of $L(t)$. However, the critical points of $L(t)$ are just the solutions of 
$$
D(t)=\frac \alpha t-\frac\beta{t+w}-\frac\gamma{t+z}=0\,, 
$$
i.e., the roots of the quadratic polynomial $(\alpha-\beta-\gamma)t^2+[\text{some junk}]t+\alpha zw$. Since the leading coefficient is negative and the free term positive, there are two roots whose product is negative none of which lies on the imaginary axis (because $\Re D(t)<0$ for imaginary $t\ne 0$), so the right half-plane contains exactly one root, which means that $r\_\*$ and the associated saddle point $t\_\*=t(r\_\*)$ are unique.


Now we are ready to define $\Gamma$. It is just the union of two steepest descents for $L(t)$ originating at $t\_\*$ one to $\infty$ and one to $0$. Since the boundary of the right half-plane is repelling for the descending gradient flow of $L$ except at the origin (it is the same condition $\Re D(t)<0$ for imaginary $t\ne 0$; the vector field in question is just $-\bar D(t)$), we are guaranteed to stay in the right half-plane all the way. The descents cannot cross $S\_{r\_\*}$ because the values of $L$ on them are smaller than $L(t\_\*)=\min\_{S\_{r\_\*}}L(t)$, and $L$ goes down at fixed rate on any compact subset of $\mathbb C\_+\setminus\{0,t\_\*,\infty\}$, so the only thing the descents can do is to escape to $0$ inside the circle and to $\infty$
outside. Near $0$ and $\infty$ the flow is asymptotically like a straight line (the direction of the field is quite clear there), so we, indeed, get an admissible contour analytic everywhere except, perhaps, $t\_\*$. Finally, the argument of $F$ stays fixed because the gradient curves of a harmonic function $L=\Re\log F$ are exactly the level curves of its harmonic conjugate $\arg F=\Im\log F$.


As I said, if we had the circle crossing property, we would be done by now. Alas, we (or at least I) don't know how to prove that, so we'll have to consider the possibility of multiple crossings. 


In the multiple crossing case the formula gets a bit more complicated:
$$
\Re\left[\int\_\Gamma e^{-i\theta}F(t)\frac{dt}t\right]=\int\_0^\infty \left[\sum\_{t\in\Gamma\cap S\_r}\varepsilon(t)|F(t)|\right]d(\log r)
$$
where $\varepsilon(t)$ is $+1$ if $\Gamma$ (oriented from $0$ to $\infty$) crosses $S\_r$ at $t$ in the outward direction and $-1$ if it does so in the inward direction.


What saves the day is that the outward and the inward crossings alternate on $S\_r$ (note that it is by no means necessary that the adjacent crossings on $S\_r$ are adjacent on $\Gamma$; still the alternation is always there because if we complement $\Gamma$ by a piece of the imaginary line and close the contour near infinity by a circular arc, we'll get a Jourdan curve bounding a domain and the signs at the crossings will correspond to going in or out of that domain when moving along the semi-circle and traversing the boundary of the domain). Now we have the following simple 


**Lemma:** Assume that a function $f(s)$ has a unique minimum on $[-1,1]$ and is monotonically increasing if we go from that minimum to any endpoint. Then for any $-1\le s\_0<s\_1<\dots<s\_{2m}\le 1$, we have
$$
\sum\_{j=0}^{2m}(-1)^jf(s\_j)\ge\min\_j f(s\_j)\ge\min\_{s\in[-1,1]}f(s)
$$
Applying this lemma to $f(\sin\theta)=|F(re^{i\theta})|, \theta\in[-\frac\pi 2,\frac\pi 2]$ finishes the story. 


To prove the lemma just consider two cases. First, suppose that all points are on the same side of the minimum. By symmetry, WLOG, we may assume that they are to the right. Then $f(s\_{2k})-f(s\_{2k-1})\ge 0$ for $k=1,\dots,m$, so the sum is at least $f(s\_0)$. Assume now that the minimum is between $s\_j$ and $s\_{j+1}$. Again, by symmetry, we may assume that $j=2j'-1$. Then $f(2k)-f(2k+1)\ge 0$ for $k=0,\dots,j'-1$ and $f(2k)-f(2k-1)\ge 0$ for $k=j'+1,\dots,m$, so the sum is at least $f(2j')=f(j+1)$.


**The End!**


Note that it proves that for $c\ge a,b\ge 0$, we have $|{}\_2F\_1(a,b;c,1-z)|\ge {}\_2F\_1(a,b;c,1-r)>0$ in $|z|<r$ for the principal branch. In particular, the principal brunch has no zeroes. The literature devoted to zeroes of ${}\_2F\_1$ is quite extensive, so such simple fact should certainly be known and its proof may shed some light on your question too (however I leave the *really difficult* task of searching the literature to you or somebody else). 


Also, if you are the same person as Roger in [Conjectured bound on Kummer's function (confluent hypergeometric function)](https://mathoverflow.net/questions/140151), I suggest that you merge the accounts. Multiple accounts are not a strict no-no but they are generally frowned upon and impede the communication. If not, just ignore this side remark :-)


