Q:

How do we prove that $\max\{x_1 + x_2+ \ldots + x_n - n + 1,0\} \leq C(\textbf{x}) \leq \min\{x_1,x_2,\ldots,x_n\}$? 

I am interested in proving the generalized version of the Fréchet-Hoeffding inequality.  Precisely speaking, given a $n$-copula $C:[0,1]^{n}\rightarrow[0,1]$, how do we demonstrate that
$$
\max\{x_1 + x_2 + \ldots + x_n - n + 1, 0\} \leq C(\textbf{x}) \leq \min\{x_1,x_2,\ldots,x_n\}
$$
MY ATTEMPT
Since $\textbf{x} = (x_1,x_2,\ldots,x_n) \leq (1,1,\ldots,1)$, I have been able to prove the upper bound inequality as next
\begin{align*}
C(\textbf{x}) & \leq C(x_1,x_2,\ldots,x_{n-1},1)\\
& \leq C(x_1,x_2,\ldots,1,1) \leq \ldots\\
& \leq C(x_1,1,\ldots,1,1) = x_1
\end{align*}
because copulas are non-decreasing in each argument and have uniform margins.  Once the same reasoning applies to each coordinate, the result $C(\textbf{x}) \leq \min\{x_1,x_2,\ldots,x_n\}$ follows. 
But what about the first inequality?  Any help is appreciated. 

A:

$C(\mathrm x)$ is a cdf of a vector $(U_1,\dots,U_n)$ with $U(0,1)$ marginals.  That said, for any $\mathrm x = (x_1,\dots x_n)\in [0,1]^n$, 
$$
C(\mathrm x) = \mathrm{P} (U_1\le x_1,\dots, U_n\le x_n)  = 1 - \mathrm{P} \left( \bigcup_{i=1}^n \{U_i> x_i\}\right)\\
\ge 1 - \sum_{i=1}^n \mathrm{P} (U_i> x_i) = 1 - \sum_{i=1}^n (1- x_i) = \sum_{i=1}^n x_i - n+ 1. 
$$

