Is $\frac{(\alpha)^n (\beta)^n} {(\delta)^n} > \frac{(\alpha+1)^n (\beta+1)^n} {(\delta+1)^n}$ for any $n$ ?(in this specific case)

Question

Let $\alpha=(K-1)a$, $\beta=K$ and $\delta=Ka$, where $K>a\ge 1$ ($\delta>\alpha>\beta$).

Can we claim that $\frac{(\alpha)^n (\beta)^n} {(\delta)^n} > \frac{(\alpha+1)^n (\beta+1)^n} {(\delta+1)^n}$ for any $n$ ? In other words : is $ _2F_1(\alpha,\beta,\delta,x) >$ $ _2F_1(\alpha+1,\beta+1,\delta+1,x)$ ?

Here $(.)^n$ is the (rising) Pochhammer symbol.

Answer 1

The ratio of $\frac{(\alpha)^n (\beta)^n} {(\delta)^n}$ to $\frac{(\alpha+1)^n (\beta+1)^n} {(\delta+1)^n}$ is exactly $$ \frac{\alpha \beta (\delta+n)}{(\alpha+n)(\beta+n)\delta}. $$ This is less than $1$ precisely when $\alpha \beta (\delta+n) < (\alpha+n)(\beta+n)\delta$, which is equivalent to $\alpha\beta < \delta(n+\alpha+\beta)$. Under the assumptions of the problem, this is always true. Therefore $\frac{(\alpha)^n (\beta)^n} {(\delta)^n} < \frac{(\alpha+1)^n (\beta+1)^n} {(\delta+1)^n}$ always.