Identity with sums of binomials Choose any natural numbers  $a \ge 1$ and $b \ge 1$. Then prove that the identity holds:
\begin{equation*}
\frac{a b }{(a+b)^2} {\Large(}\tbinom{a+b}{a} {\Large)}^2   = \\
\sum_{n=1}^{a}\sum_{m=1}^{b}  {\Huge[}   {\Large(}\tbinom{n+m-2}{n-1} {\Large)}^2 - \frac{ m (a-n)(b-m)  }{n(a-n+b-m)^2} {\Large(}\tbinom{n+m-1}{n-1} {\Large)}^2  {\Huge]} {\Large(}\tbinom{a-n+b-m}{a-n} {\Large)}^2 
\end{equation*}
In here, the fraction $\frac{ (a-n)(b-m)  }{(a-n+b-m)^2}$ is understood to be zero for $a=n$ and $b=m$.
Alternative:
Show that the following equation is in fact an identity for arbitrary $x,y$ with $x+y<1$. 
\begin{equation*}
1 = {\Huge[} 1 + \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} {x^n y^m} \frac{2 m n }{(m+n)^2} {\Large(}\tbinom{n+m}{n} {\Large)}^2 {\Huge]}^2 - 4 x y {\Huge[} \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} {x^n y^m} {\Large(}\tbinom{n+m}{n} {\Large)}^2  {\Huge]}^2
\end{equation*}
Indeed, the top identity is - after some manipulation - an equation for the coefficient of $x^a y^b$ in the latter identity.
My effort: I tried to simplify the binomials further but had no success in summing up. A computer program verifies (up to reasonable numbers $a$ and $b$) that indeed equality holds.  

The answer can be derived from two answers to this question: There, the topic is that the partial derivatives of 
$f(x,y)=(1+x^2+y^2-2x-2y-2xy)^{-\frac{1}{2}}$
satisfy
\begin{equation*}
\frac{\partial ^{m+n}}{\partial x^{m} \partial y^{n}} f (0,0)= (n+m)! \tbinom{n+m}{n}
\end{equation*}
Now this relation is shown directly (second answer to the said question), and then, by a Taylor expansion of $f(x,y)$, also the following equation holds:
\begin{equation*}
1 = (1+x^2+y^2-2x-2y-2xy) \cdot {\Huge[} \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} {x^n y^m} {\Large(}\tbinom{n+m}{n} {\Large)}^2  {\Huge]}^2
\end{equation*}
This equation, after applying $1+x^2+y^2-2x-2y-2xy = (1-x-y)^2 - 4xy$, equals 
\begin{equation*}
1 = {\Huge[} 1 + \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} {x^n y^m} \frac{2 m n }{(m+n)^2} {\Large(}\tbinom{n+m}{n} {\Large)}^2 {\Huge]}^2 - 4 x y {\Huge[} \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} {x^n y^m} {\Large(}\tbinom{n+m}{n} {\Large)}^2  {\Huge]}^2
\end{equation*}
which establishes the  identity required here.
