Showing the Clayton Copula is $2-$increasing This is the definition of a bivariate ($2$-dimensional) copula:

$C(\mathbf{u}):[0,1]^2 \mapsto [0,1]$ is a bivariate copula if
  
  
*
  
*$C(u_{1},0) = 0$ and $C(0,u_{2})=0$; i.e., $C = 0$ if one argument is $0$.
  
*$C(u_{1},1) = u_{1}$ and $C(1,u_{2}) = u_{2}$; i.e., the copula reduces to $u_{i}$ if all arguments are $1$ except the $i$th one.
  
*$C(\mathbf{u})$ is $2$-increasing - i.e., for each hyperrectangle $B = \prod_{i=1}^{k}[x_{i},y_{i}]$ in $[0,1]^{2}$, the $C$-volume:
  $$ \int_{B}dC = \sum_{\mathbf{z} \in \{x_{1},y_{1}\}\times\{x_{2},y_{2}\}} (-1)^{N(\mathbf{z})} C(\mathbf{z}) \geq 0 $$
  where $N(\mathbf{z}) = \text{the number of}\,z_{i}=x_{i}$ for $\mathbf{z} \in \{x_{1},y_{1}\}\times \{x_{2},y_{2}\}$

I need to prove that the Clayton Copula, $C(u,v) = \left[\max\{u^{-\theta} + v^{-\theta}-1,0 \}\right]^{-1/\theta}$ for $u,v \in (0,1)$ and $\theta > 0$, is a bonafide bivariate copula.
So, far, the only part I am still having trouble with is showing property #3 - namely, that $C$ is what is known as $2-$increasing.
For #3, I have that $\displaystyle \int_{B}dC = C(x_{2},y_{2})-C(x_{2},y_{1})-C(x_{1},y_{2})+C(x_{1},y_{1})$, which after many, many steps of algebra, I got to look like $$ = \frac{x_{2}y_{2}}{\left(y_{2}^{\theta}+x_{2}^{\theta}-x_{2}^{\theta}y_{2}^{\theta} \right)^{1/\theta}} - \frac{x_{2}y_{1}}{\left( y_{1}^{\theta}+x_{2}^{\theta}-x_{2}^{\theta}y_{1}^{\theta} \right)^{1/\theta}} - \frac{x_{1}y_{2}}{\left(y_{2}^{\theta}+x_{1}^{\theta}-x_{1}^{\theta}y_{2}^{\theta} \right)^{1/\theta}}  + \frac{x_{1}y_{1}}{\left( y_{1}^{\theta}+x_{1}^{\theta}-x_{1}^{\theta}y_{1}^{\theta} \right)^{1/\theta}} $$
But, how do I show that this must be $\geq 0$?
I thank you ahead of time for your help!
 A: There may be some trick to do this directly, but there is another way if you are familiar with what is called the generator of the copula.
Note that for $\theta > 0$ and $u,v \in (0,1)$ we can drop the $\max$ in the definition of the Clayton copula since $u^{-\theta} + v^{-\theta} - 1 > 0$.

Theorem: Let $\phi :[0,1] \to [0,\infty)$ be a continuous, non-increasing function where $\phi(1) = 0$.  There is a more general
  definition of a generator, but if $\phi(t) \to \infty$ as $t \to 0$,
  then 
$$C(u,v) = \phi^{-1}(\phi(u) + \phi(v))$$ 
satisfies the boundary properties of a copula and if $\phi$ is
  convex then $C$ is $2-$increasing and, hence, a copula.

Consider $\phi(t) = (t^{-\theta}-1)/\theta$. Note that $\phi$ is non-increasing, $\phi(1) = 0$ and $\phi(0) = +\infty$.  Also, $\phi''(t) > 0$ and, therefore, $\phi$ is convex.
We can show that this function $\phi$ generates the Clayton copula. 
Observe that 
$y = \phi(t) = (t^{-\theta}-1)/\theta $   implies that $t = (1 +\theta y)^{-1/\theta}$, and we see that the inverse function is $\phi^{-1}(t) = (1 + \theta t)^{-1/\theta}$.
Hence,
$$C(u,v) = \left[1 + \theta \left(\frac{u^{-\theta}-1}{\theta} + \frac{v^{-\theta}-1}{\theta}\right)\right]^{-1/\theta} = (u^{-\theta} + v^{-\theta} - 1)^{-1/\theta}$$
