Here $n,N$ are non-negative integers, $2n+1\leq N$, and $I$ is the regularized Beta function, $I\left(x,n,N-n\right)=\frac{B\left(x,n,N-n\right)}{B\left(n,N-n\right)}=\frac{1}{B\left(n,N-n\right)}\int_{0}^{x}t^{n-1}\left(1-t\right)^{N-n-1}dt$.
Context
Consider we have a set of $N-1$ uniformly distributed random numbers within $\left[0,1\right]$ and we are adding $x\sim U\left[0,1\right]$ to the set, which ended up to have $N$ elements.
Order statistics tell us that $I\left(x,n,N-n\right)-I\left(x,n,N-n\right)$ is the probability of having the $x$ to be no less than the $n$th smallest number and no greater than the $n$th largest number. Meanwhile, $1-2n/N$ is the normalization factor, which is the average probability of having $x$ falls within the middle $1-2n/N$ fraction of the distribution across all $x\in\left[0,1\right]$.
Notice we have $\int_{0}^{1}\frac{I\left(x,n,N-n\right)-I\left(x,n,N-n\right)}{1-2n/N}dx=1$, which confirms that $1-2n/N$ is the normalization factor. Moreover, the Holder's inequality suggests that $$ \int_{0}^{1}\left[\frac{I\left(x,n,N-n\right)-I\left(x,n,N-n\right)}{1-2n/N}\right]^{2}dx \geq \left(\int_{0}^{1}\frac{I\left(x,n,N-n\right)-I\left(x,n,N-n\right)}{1-2n/N}dx\right)^2=1. $$
My intuition is that, if $\frac{I\left(x,n,N-n\right)-I\left(x,n,N-n\right)}{1-2n/N}$ has less ``variation'' across $x\in\left[0,1\right]$, the integral of square gets closer to the lower bound $1$. At the same time, a smaller $n$ with respect to $N$ means the probability on the numerator converges to $1-2n/N$. Specifically, when $N$ is very large and $n$ is very small, we have both the numerator $I\left(x,n,N-n\right)-I\left(x,n,N-n\right)$ and denominator $1-2n/N$ get very close to $1$ for all $x$.
Hence, I have the conjecture that $\int_{0}^{1}\left[\frac{I\left(x,n,N-n\right)-I\left(x,n,N-n\right)}{1-2n/N}\right]^{2}dx$ increasing in $n$, which has been confirmed under quite a few numerical cases. But I have a difficult time proving this. Thanks!