Inequality involving Pochhammer symbols Let $m,S$ be integers satisfying $2\leq m\leq S$. I would like to show that
$$h_1\left(x\right) h_3\left(x\right) \leq h_2^2\left(x\right)$$
for all $x\geq 0$ where
$$h_k\left(x\right) \equiv {}_2F_1\left(k,k-1-m;S-m+k,-x\right).$$
${}_2F_1$ denotes the hypergeometric function.

What I have so far...
Note that \begin{multline*}
h_{1}\left(y\right)h_{3}\left(y\right)=\sum_{d=0}^{\infty}\left(-y\right)^{d}\sum_{p=0}^{d}\frac{\left(1\right)^{\left(p\right)}\left(-m\right)^{\left(p\right)}}{\left(S-m+1\right)^{\left(p\right)}p!}\frac{\left(3\right)^{\left(q\right)}\left(2-m\right)^{\left(q\right)}}{\left(S-m+3\right)^{\left(q\right)}q!}\\
=\sum_{d=0}^{\infty}y^{d}\sum_{p=0}^{d}\frac{m_{p}}{\left(S-m+1\right)^{\left(p\right)}}\frac{\left(q+1\right)\left(q+2\right)\left(m-2\right)_{q}}{2\left(S-m+3\right)^{\left(q\right)}}
\end{multline*} where $q\equiv d-p$ and the Pochhammer symbols are
  defined here: https://en.wikipedia.org/wiki/Pochhammer_symbol.
  Similarly, \begin{multline*}
h_{2}^{2}\left(y\right)=\sum_{d=0}^{\infty}\left(-y\right)^{d}\sum_{p=0}^{d}\frac{\left(2\right)^{\left(p\right)}\left(1-m\right)^{\left(p\right)}}{\left(S-m+2\right)^{\left(p\right)}p!}\frac{\left(2\right)^{\left(q\right)}\left(1-m\right)^{\left(q\right)}}{\left(S-m+2\right)^{\left(q\right)}q!}\\
=\sum_{d=0}^{\infty}y^{d}\sum_{p=0}^{d}\frac{\left(p+1\right)\left(m-1\right)_{p}}{\left(S-m+2\right)^{\left(p\right)}}\frac{\left(q+1\right)\left(m-1\right)_{q}}{\left(S-m+2\right)^{\left(q\right)}}.
\end{multline*}
  So it is sufficient to show $$
\sum_{p=0}^{d}\frac{\left(q+1\right)\left(q+2\right)m_{p}\left(m-2\right)_{q}}{2\left(S-m+1\right)^{\left(p\right)}\left(S-m+3\right)^{\left(q\right)}}-\frac{\left(q+1\right)\left(p+1\right)\left(m-1\right)_{p}\left(m-1\right)_{q}}{\left(S-m+2\right)^{\left(p\right)}\left(S-m+2\right)^{\left(q\right)}}\leq0, $$ for all $d\geq0$.
Note that for $d \geq 2m - 1$, the above is zero (see also
  @Semiclassical's comment below), so that we only have to verify the
  claim for $0 \leq d \leq 2\left(m -1\right)$.

 A: I'll prove here some other easy cases, the first being in the comments.
My numerical experiments indicate that one should require $2 \le k \le m \le S$ to show
\begin{align}
q_k(x) := h_{k-1}(x)h_{k+1}(x) - h_{k}^2(x) \le 0
\end{align}
Step 0, one sees immediately that $h_k(x) = h_{k,S,m}(x)$ is a polynomial of degree $(m-k+1)$.
Step 1, $m=k$, one has
\begin{align}
h_{k-1,S,k}(x) &= 1+\frac{(2k-2)}{(S-1)}x+\frac{k(k-1)}{S(S-1)}x^2\\
h_{k,S,k}(x) &= 1+\frac{k}{S}x\\
h_{k+1,S,k}(x) &= 1
\end{align}
which gives
\begin{align}
q_k(x)
&= h_{k-1,S,k}(x)\:h_{k+1,S,k}(x)-h_{k,S,k}(x)^2\\
&= \left(1+\frac{(2k-2)}{(S-1)}x+\frac{k(k-1)}{S(S-1)}x^2\right)\cdot 1 -\left(1+\frac{k}{S}x\right)^2\\
&= -x(2S+kx)\frac{(S-k)}{(S-1)S^2}.
\end{align}
This is clearly a degree 2 polynomial in x with zeros $\{-2S/k,0\}$.
So $x\ge0$ gives $q_k(x) \le 0$. $\square$
Step 2, $m=k+1$, $2 \le k \le k+1 \le S$, we have
\begin{equation}
(5kS+3S-k-3)\ge (5k(k+1)+3(k+1)-k-3) = 5k^2+7k+6 > 0,
\end{equation}
which gives
\begin{align}
q_k(x) = -x(2S(S^2-1)+S(5kS+3S-k-3)x
+4k(k+1)Sx^2+k(k+1)^2x^3)\frac{(S-k-1)}{(S-2) (S-1)^2 S^2} \le 0. \square
\end{align}
Now the easy coefficients of the general case. Step 0, gives that $\text{deg}(q_k(x)) = 2m-2k+2$.
Step 3, $r\ge 2$, $m=k+r$, $2\le k\le k+r \le S$, one has, without proving the equality,
\begin{align}
[x^0](q_k(x)) &= 0,\\
[x^1](q_k(x)) &= -2\cdot\frac{(S+1)(S-k-r)}{(S-r-1)(S-r+0)(S-r+1)} \le 0,\\
[x^{2r+1}](q_k(x)) &= -(2r+2)\cdot\frac{
(S-k-r)
\prod\limits_{\nu=0}^{r-1}(k+\nu)
\prod\limits_{\nu=1}^{r}(k+\nu)
}{
\prod\limits_{\nu=0}^{r}(S-\nu)
\prod\limits_{\nu=1}^{r+1}(S-\nu)
} \le 0,\\
[x^{2r+2}](q_k(x)) &= -\frac{
(S-k-r)
\prod\limits_{\nu=0}^{r}(k+\nu)
\prod\limits_{\nu=1}^{r}(k+\nu)
}{
\prod\limits_{\nu=0}^{r}(S-\nu)
\prod\limits_{\nu=0}^{r+1}(S-\nu)
} \le 0.\square
\end{align}
Next the remaining three cases for $r=2$.
Step 4, $r=2$, $m=k+r$, $2\le k\le k+r \le S$, we have
\begin{align}
(3kS+2S-4k-4) & \ge (3k(k+2)+2(k+2)-4k-4) \ge 3k^2+4k > 0,\\
(4kS+2S+k-4) & \ge (4k(k+2)+2(k+2)+k-4) \ge 4k^2+11k > 0, \\
(7kS+11S-2k-7) & \ge (7k(k+2)+11(k+2)-2k-7) \ge 7k^2+23k+15 > 0,
\end{align}
which gives, without proving the equality,
\begin{align}
[x^2](q_k(x)) &= -3(3kS+2S-4k-4)\cdot\frac{(S-k-2)(S+1)}{(S-3)(S-2)^2(S-1)S} \le 0,\\
[x^3](q_k(x)) &= -4(4kS+2S+k-4)\cdot\frac{(S-k-2)(k+1)}{(S-3)(S-2)^2(S-1)S}
\le 0,\\
[x^4](q_k(x)) &= -2(7kS+11S-2k-7)\cdot\frac{(S-k-2)k(k+1)}{(S-3)(S-2)^2(S-1)^2S} \le 0. \square
\end{align}
