Lower bound on hypergeometric function ${}_2F_1$ I am at my wit's end trying to show the following lower bound:
$${}_2F_1[-m, -m; -(m+l); z]\geq (1-z)^m,\tag{1}$$
where ${}_2F_1[a,b;c;z]$ is the Gauss hypergeometric function, $m,l=0,1,2,\ldots$, and $0<z<1$.  Numerical experiments seem to confirm that this bound holds, but I cannot figure out how to prove it. Any help?
What I tried
When argument $c$ in ${}_2F_1[a,b;c;z]$ is a negative integer ${}_2F_1[a,b;c;z]$ is usually undefined, however, per discussion on DLMF, here we can express the LHS as
$${}_2F_1[-m, -m; -(m+l); z]=\sum_{n=0}^m (-1)^n\binom{m}{n}\frac{(-m)_n}{(-(m+l))_n}z^n=\sum_{n=0}^m \binom{m}{n}\frac{\binom{m}{n}}{\binom{m+l}{n}}(-z)^n,$$
where $(-m)_n=\left\{\begin{array}{rl}\frac{(-1)^nm!}{(m-n)!},&0\leq n\leq m  \\ 0, &n>m\end{array}\right.$ is the Pochhammer's symbol for $m$ and $n$ nonnegative integers.
To prove the bound above, I originally attempted to show that
$$\binom{m}{n}\frac{\binom{m}{n}}{\binom{m+l}{n}}z^n-\binom{m}{n+1}\frac{\binom{m}{n+1}}{\binom{m+l}{n+1}}z^{n+1}\geq\binom{m}{n}z^n-\binom{m}{n+1}z^{n+1}\tag{2}$$
for $m,l,n=0,1,2,\ldots$ and $0<z<1$.
Edit: @VarunVejalla pointed to a counterexample $m,l,n=2,2,1$ in a comment, showing that (2) doesn't hold.  However, this doesn't rule out the truth of (1).
Any ideas how to prove (1), or a counterexample to it, are appreciated!
 A: Proof by induction:
Inequality $(1)$ holds for $l = 0$
$$\begin{align*}{}_2F_1[-m, -m; -(m+0); z]&=\sum_{n=0}^m (-1)^n\binom{m}{n}\frac{(-m)_n}{(-(m+0))_n}z^n\\
\\
&=\sum_{n=0}^m \binom{m}{n}\left(-z\right)^n\\
\\
&=(1-z)^m \quad \text{(binomial theorem)}\\
\\
\therefore \;{}_2F_1[-m, -m; -(m+0); z]&\ge  (1-z)^m \\
\end{align*}$$
Assume inequality $(1)$ holds for $l-1$, namely:
$${}_2F_1[-m, -m; -(m+l-1); z]\ge  (1-z)^m$$
So now
$$\begin{align*} {}_2F_1[-m, -m; -(m+l); z] &=\sum_{n=0}^m (-1)^n\binom{m}{n}\frac{(-m)_n}{(-(m+l))_n}z^n\\
\\
&= \sum_{n=0}^m (-1)^n\binom{m}{n}\frac{(-m)_n}{\left(\frac{(-1)^n(m+l)!}{(m+l-n)!}\right)}z^n\\
\\
&= \sum_{n=0}^m (-1)^n\binom{m}{n}\frac{(-m)_n}{\frac{(-1)^n(m+l-1)!}{(m+l-1-n)!}\cdot\frac{m+l}{m+l-n}}z^n\\
\\
&= \sum_{n=0}^m \dfrac{m+l-n}{m+l}(-1)^n\binom{m}{n}\frac{(-m)_n}{(-(m+l-1))_n}z^n\\
\\
&= {}_2F_1[-m, -m; -(m+l-1); z] \\
&\quad + \dfrac{1}{m+l}\sum_{n=0}^m n(-1)^{n+1}\binom{m}{n}\frac{(-m)_n}{(-(m+l-1))_n}z^n \\
\\
&= {}_2F_1[-m, -m; -(m+l-1); z] \\
&\quad - \dfrac{z}{m+l}\sum_{n=1}^m \frac{(-m)_n(-m)_n}{(-(m+l-1))_n}\dfrac{z^{n-1}}{(n-1)!} \\
\\
&=  {}_2F_1[-m, -m; -(m+l-1); z] \\
&\quad - \dfrac{z}{m+l}\sum_{k=0}^{m-1} \frac{(-m)_{k+1}(-m)_{k+1}}{(-(m+l-1))_{k+1}}\dfrac{z^{k}}{k!} \\
\\
&= {}_2F_1[-m, -m; -(m+l-1); z] \\
&\quad + \dfrac{m^2 z}{(m+l)(m+l-1)}\sum_{k=0}^{m-1} \frac{(-(m-1))_{k}(-(m-1))_{k}}{(-(m-1+l-1))_{k}}\dfrac{z^{k}}{k!} \\
\\
&= {}_2F_1[-m, -m; -(m+l-1); z] \\
&\quad + \dfrac{m^2 z}{(m+l)(m+l-1)}{}_2F_1[-(m-1), -(m-1); -(m-1+l-1); z] \\
\\
{}_2F_1[-m, -m; -(m+l); z] &\ge (1-z)^m + \dfrac{m^2 z}{(m+l)(m+l-1)}(1-z)^{m-1}\\
\\
\therefore \; {}_2F_1[-m, -m; -(m+l); z] &\ge (1-z)^m\\
\end{align*}$$
Whew!
Please check all the steps.  I make no guarantees that the above is free from errors. :)
