Let $\mu_i,\nu_i$ be probability measure on a finite space $\Omega_i,i=1,2,\dots,n$. Define $\mu=\prod\limits_{i=1}^{n}\mu_i$ and $\nu=\prod\limits_{i=1}^{n}\nu_i$ on $\Omega=\prod\limits_{i=1}^{n}\Omega_i$, show that $\|\mu-\nu\| \le \sum\limits_{i=1}^{n}\|\mu_i-\nu_i\|$ where $\|\mu-\nu\|$ denote the total variation distance between $\mu$ and $\nu$.
I know how to do this using coupling, is there a way to do it without coupling?
I try to write $\|\mu-\nu\|={1 \over 2}\sum\limits_{x=(x_1,x_2,\dots,x_n) \in \Omega}|\prod\limits_{i=1}^{n}\mu_i(x_i)-\prod\limits_{i=1}^{n}\nu_i(x_i)|$
and use the fact that $\prod\limits_{i=1}^{n}\mu_i \le \sum\limits_{i=1}^{n}\mu_i$ and $\prod\limits_{i=1}^{n}\nu_i \le \sum\limits_{i=1}^{n}\nu_i$, but I didn't succeed.