Inequality involving sums of fractions of products of binomial coefficients Let $n\in\mathbb{N}$.
For $0\le l\le n$ consider
\begin{equation}
 b_l:=4^{-l} \sum_{j=0}^l \frac{\binom{2 l}{2 j} \binom{n}{j}^2}{\binom{2 n}{2 j}}\text{.} 
\end{equation}
Do you know a technique how to prove that
\begin{equation}
 b_l\ge b_n\text{,$\quad 0\le l\le n-1$?}
\end{equation}
Going through a long list of binomial identities I did not find epiphany.
Addition: Plot of $b_l$ for $n=20$.

 A: The sum $b_\ell$ is clearly hypergeometric:
$$
  b_\ell = 4^{-\ell} \sum_{j=0}^\ell \frac{\binom{2\ell}{2 j} \binom{n}{j}^2}{\binom{2n}{2j}} = 4^{-\ell} \sum_{j=0}^\ell \frac{(-n)_j}{\left(\frac{1}{2}-n\right)_j} \frac{(-\ell)_j \left(\frac{1}{2}-\ell\right)_j}{j! \cdot j!} =  4^{-\ell} {}_3F_2\left(\left. \begin{array}{ccc} -\ell & \frac{1}{2} -\ell & -n \\ & 1 & \frac{1}{2}-n \end{array} \right| 1\right)
$$
This representation allows to find
$$
   b_n = 4^{-n} {}_3F_2\left(\left. \begin{array}{ccc} -n & \frac{1}{2} -n & -n \\ & 1 & \frac{1}{2}-n \end{array} \right| 1\right) = 4^{-n} {}_2F_1\left(\left. \begin{array}{cc} -n &  -n \\ & 1  \end{array} \right| 1\right) = \frac{1}{4^n} \binom{2n}{n}
$$
The above function allows to extend the sequence to $\ell > n$. This sequence is not decreasing for all $\ell \geqslant 0$, but does appear to decrease on the interval $0 \leqslant \ell\leqslant n$. Here is an example for $n=20$:


Now, to the probabilistic interpretation of the $b_\ell$. Suppose an urn contains $2n$ balls, $n$ white and $n$ blue. We sample $m$ balls without the replacement. The probability that the sample contains equal number of balls of different colors is
$$
    p_m = \cases{\frac{\binom{n}{j} \binom{n}{j}}{\binom{2n}{2j}} & $m=2j$ \\ 0 & $m = 2j+1$}
$$
If the size of the sample follows a symmetric binomial distribution, the probability of getting sample with equal number of colors is:
$$
   b_\ell = \sum_{j=0}^\ell \frac{\binom{n}{j} \binom{n}{j}}{\binom{2n}{2j}} \binom{2\ell}{2j} 4^{\ell}
$$
I am not seeing how to establish the inequality though, but hope this helps.
