Minimum of $\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} $. $\forall n \in \mathbf{N}$, prove that the function $f(x)=\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} $ attains its minimum at $x=\frac{1}{2}$.
Now it suffices to prove that
\begin{equation*}
\sum_{l=0}^{n} \binom{n}{l}^2 (x+y)^{2l} (x-y)^{2(n-l)} = \sum_{l=0}^{n} \binom{2l}{l} \binom{2(n-l)}{n-l} x^{2l}y^{2(n-l)}
\end{equation*}

\begin{align*}
n &> 0  \text{.}  \\
f'(1/2) &= \left. \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l)x^{2l-1}(1-x)^{2(n-l)} - 2(n-l)x^{2l}(1-x)^{2(n-l)-1} \right) \right|_{x = 1/2}  \\
    &= \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l)(1/2)^{2l-1}(1/2)^{2(n-l)} - 2(n-l)(1/2)^{2l}(1/2)^{2(n-l)-1}  \right)  \\
    &= (1/2)^{2n-1}\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l) - 2(n-l) \right)  \\
    &= 0  \text{.}  \\
f(x) &= \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)}  \\
    &= \frac{(1-x)^{2n}}{(n!)^2} + \frac{x^{2n}}{(n!)^2} + \sum_{l=1}^{n-1} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} \text{.}  \\
f''(x) &= \frac{2n(2n-1)\left(x^{2(n-1)} + (1-x)^{2(n-1)}\right)}{(n!)^2} + \\
     {}&+ \sum_{l=1}^{n-1} \frac{1}{(l!)^2((n-l)!)^2} x^{2(l-1)}(1-x)^{2(n-l-1)} \left( 2n(2n-1)x^2 - 4l(2n-1)x + 2l(2l-1)\right)  \\
    &> 0 + \sum_{l=1}^{n-1} \frac{1}{(l!)^2((n-l)!)^2} x^{2(l-1)}(1-x)^{2(n-l-1)} \frac{2l(l-n)}{n}  \\
    &\geq 0  \text{.}  \blacksquare
\end{align*}
