Upper bound on hypergeometric function ${}_2F_1$ I wonder if the following upper bound holds:
$${}_2F_1[-m, -m; -(m+l); z]\leq 1,\tag{1}$$
where ${}_2F_1[a,b;c;z]$ is the Gauss hypergeometric function, $m,l=0,1,2,\ldots$, and $0<z<1$.
Clearly, this holds for $l=0$ since (1) then reduces to the binomial identity in the answer to a related question.  Thus, induction seems to be a promising path to proving this upper bound.  However, direct application of the expansion:
$$\begin{align}{}_2F_1[-m, -m; -(m+l); z] &= {}_2F_1[-m, -m; -(m+l-1); z] \\
&\quad + \dfrac{m^2 z}{(m+l)(m+l-1)}{}_2F_1[-(m-1), -(m-1); -(m-1+l-1); z]\end{align}$$
derived for the aforementioned solution does not seem to work here.
Numerical experiments seem to confirm this bound, though perhaps there is a counterexample. I tried induction on $m$ instead of $l$ to no avail.  Any ideas?
 A: First show that, for $m >0$, the function is always decreasing over the interval $0<z<1$:
$$\begin{align*}\dfrac{d}{dz}{}_2F_1(-m,-m;-[m+l];z) &= \dfrac{d}{dz}\sum_{n=0}^{m} \dfrac{(-m)_{n}(-m)_{n}}{(-(m+l))_{n}}\dfrac{z^{n}}{n!}\\
\\
&= \sum_{n=1}^{m} \dfrac{(-m)_{n}(-m)_{n}}{(-(m+l))_{n}}\dfrac{z^{n-1}}{(n-1)!} \\
\\
&= \sum_{k=0}^{m-1} \dfrac{(-m)_{k+1}(-m)_{k+1}}{(-(m+l))_{k+1}}\dfrac{z^{k}}{k!} \\
&= -\dfrac{m^2}{m+l} \sum_{k=0}^{m-1} \dfrac{(-(m-1))_{k}(-(m-1))_{k}}{(-(m-1+l))_{k}}\dfrac{z^{k}}{k!}\\
\\
&=  -\dfrac{m^2}{m+l} {}_2F_1(-[m-1],-[m-1];-[m-1+l];z) \\
\\
&\le -\dfrac{m^2}{m+l} (1-z)^{m-1} \\
\\
\therefore \; \dfrac{d}{dz}{}_2F_1(-m,-m;-[m+l];z) &\lt 0\\
\end{align*}$$
which used a lower bound for ${}_2F_1(-[m-1],-[m-1];-[m-1+l];z)$ that comes from this answer.
Then, for $m>0$, the function is at its maximum value in $0<z<1$ as $z\rightarrow0$:
$$\begin{align*}\lim_{z\to0}{}_2F_1(-m,-m;-[m+l];z) &= \lim_{z\to0} \sum_{n=0}^{m} \dfrac{(-m)_{n}(-m)_{n}}{(-(m+l))_{n}}\dfrac{z^{n}}{n!} \\
\\
&= \lim_{z\to0}\left[ 1 + \sum_{n=1}^{m} \dfrac{(-m)_{n}(-m)_{n}}{(-(m+l))_{n}}\dfrac{z^{n}}{n!}\right]\\
\\
\lim_{z\to0}{}_2F_1(-m,-m;-[m+l];z) &= 1\\
\end{align*}$$
One can then conclude:
$${}_2F_1(-m,-m;-[m+l];z) \le 1 \quad z\in (0,1), m > 0$$
For the special case of $m = 0$:
$${}_2F_1(0,0;-l;z) = 1$$
One can then finally conclude:
$${}_2F_1(-m,-m;-[m+l];z) \le 1 \quad z\in (0,1)$$
