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*}
 A: We want to minimize the function $f(x)=\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} $. 
Let $y = 2x(1-x)$. For any range of $x$, we have $y\leq 1/2$ , with $y=1/2$ achieved when $x=1/2$. 
Consider the following integral:
$$g(y)=\int_0^{2\pi}\left(1-y(1-\cos \theta)\right)^n d\theta$$
I could show that
$$g(y) = (n!)^2 f(x)$$
(see further below for proof of above)
Since $(1-\cos \theta)\in [0,2]$, $g(y)$ is a linear combination of monotonic non-increasing  functions. So, $g(y)$ monotonic non-increasing in $y$. Since the range of $y$ is $(-\infty,1/2]$, $g(y)$ is minimized at $y=1/2$. 
This means that $f(x)$ is minimized at $x=1/2$. 
Proof for $g(y) = (n!)^2 f(x)$:
To show the above, I considered a random variable $X$ that takes values in the set $\{0,1,...,n\}$ with following probability mass function.$$P[X=k]= p_k= \frac {n!}{k! (n-k)!}x^k (1-x)^{n-k}$$ Clearly, $\sum_k P[X=k] = \sum_k p_k  = (x+1-x)^n=1$ so this is a valid probability mass function. 
Note that $f(x) = \frac 1 {(n!)^2} \sum_k p_k^2$, so our interest is in $\sum_k p_k^2$. Consider the following transform: $$ \phi(\theta) = \sum_k p_k e^{j\theta k} = (x e^{j\theta} + 1-x)^n $$
This $\phi(\theta)$ is the moment generating function of $X$ evaluated at $j\theta$.
Clearly, $$\phi(\theta) \phi(-\theta) = (1-y+y\cos \theta)^n$$
Interestingly, this gives an expression for  $\sum_k p_k^2$ as follows:
$$\phi(\theta) \phi(-\theta) = \left(\sum_k p_k e^{j\theta k}\right) \left(\sum_k p_k e^{-j\theta k}\right) = \sum_k p_k^2 + 2 \sum_l \sum_{m>l} p_l p_m cos\left((m-l)\theta\right)$$
We can get rid of the pesky cos terms by integration.
$$g(y)=\int_0^{2\pi}\left(1-y + y \cos \theta\right)^n d\theta = \sum_k p_k^2 $$
So, $g(y) = (n!)^2 f(x)$ as promised.
A: \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*}
