if $a+b>c+d$ and $c cd$ show that if $a+b>c+d$ and $c<a<b<d$, then $ab > cd$.
I was wondering if there is a simple way to prove this statement or a counter example.
 A: You are given
$$a + b \gt c + d \tag{1}\label{eq1A}$$
$$c \lt a \lt b \lt d \tag{2}\label{eq2A}$$
You are asking about checking on the following being true
$$ab \gt cd \tag{3}\label{eq3A}$$
Here's a counter-example. Let $c = -10$, $a = -8$, $b = -7$ and $d = -6$. Then $a + b = -15 \gt c + d = -16$, but $ab = 56 \lt cd = 60$.
To check on when \eqref{eq3A} is true, let $a = c + e$, $b = c + f$ and $d = c + g$ where $0 \lt e \lt f \lt g$. The first condition now becomes
$$\begin{equation}\begin{aligned}
a + b & \gt c + d \\
c + e + c + f & \gt c + c + g \\
e + f & \gt g
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Also, you have
$$\begin{equation}\begin{aligned}
ab - cd & = (c + e)(c + f) - c(c + g) \\
& = c^2 + (e + f)c + ef - c^2 - cg \\
& = (e + f - g)c + ef
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
For \eqref{eq3A} to be true requires
$$(e + f - g)c + ef \gt 0 \iff c \gt -\frac{ef}{e + f - g} \tag{6}\label{eq6A}$$
Since $e + f \gt g \implies e + f - g \gt 0$, the RHS of \eqref{eq6A} is negative. Thus, for example, if all of the values are required to be non-negative, i.e., that $c \ge 0$, then \eqref{eq2A} will be true.
However, if instead $c$ is a value much less than $0$ compared to how $e$, $f$ and $g$ are greater than $0$, then $c$ could be less than or equal to the RHS of \eqref{eq6A}. This is basically how I determined my counter-example, with $c = -10$ being just a relatively arbitrary value I chose.
A: Assume $0<c<a<b<d$. From the given $a+b>c+d$, we have
$$a-c > d - b>0$$
Then, multiply $b>c>0$ to get
$$(a-c)b > (d - b)c$$
Thus, 
$$ab > cd$$
