Total variation inequality for the product measure 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.
 A: This is a direct consequence of the fact that for every nonnegative $a_i$ and $b_i$,
$$|(a_1\cdots a_n)-(b_1\cdots b_n)|\leqslant\sum\limits_{i=1}^n|a_i-b_i|\,(a_1\cdots a_{i-1})(b_{i+1}\cdots b_n).
$$
Hence,
$$
2\|\mu-\nu\|=\sum\limits_x\left|\mu_1(x_1)\cdots\mu_n(x_n)-\nu_1(x_1)\cdots\nu_n(x_n)\right|\leqslant\sum\limits_{i=1}^n\Delta_i,
$$
with
$$
\Delta_i=\sum\limits_{x_i}|\mu_i(x_i)-\nu_i(x_i)|\,\sum\limits_{\widehat x_i}\mu_1(x_1)\cdots\mu_{i-1}(x_{i-1})\nu_{i+1}(x_{i+1})\cdots\nu_n(x_n),
$$
where $\widehat x_i=(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n)$. Each sum over $\widehat x_i$ is a product of masses of probability measures hence
$$
\Delta_i=\sum\limits_{x_i}|\mu_i(x_i)-\nu_i(x_i)|=2\|\mu_i-\nu_i\|,
$$
and you are done.
Edit The first inequality is a consequence of the triangular inequality between the numbers $(c_i)_{0\leqslant i\leqslant n}$ defined by $c_i=(a_1\cdots a_{i})(b_{i+1}\cdots b_n)$ for $1\leqslant i\leqslant n-1$, $c_0=b_1\cdots b_n$ and $c_n=a_1\cdots a_n$ since, for every $1\leqslant i\leqslant n$,
$$
c_{i}-c_{i-1}=(a_{i}-b_{i})(a_1\cdots a_{i-1})(b_{i+1}\cdots b_n).
$$
A: First, notice that the difference of two probability measures is a signed measure of total variance at most $2$.
Hence your problem reduces to showing that $\|\mu\|\leq\sum\limits_{i=1}^{n}\|\mu_i\|$ for signed measures $\mu_i$ on finite spaces $\Omega_i$ with $\|\mu_i\|\leq 2$.
I will show you how to do the case $n=2$.
Then we have to show that
$$
\sum\limits_{x\in\Omega_1,y\in\Omega_2} \lvert\mu_1(x)\mu_2(y)\rvert\leq
\sum\limits_{x\in\Omega_1} \lvert\mu_1(x)\rvert
+\sum\limits_{y\in\Omega_2} \lvert\mu_2(y)\rvert.
$$
But this inequality is equivalent to
$$\|\mu_1\|\cdot\|\mu_2\|\leq\|\mu_1\|+\|\mu_2\|,$$
which is easily verified using $\|\mu_i\|\leq 2$:
Edit: The last inequality can be seen as follows:
$$
\|\mu_1\|\cdot\|\mu_2\|\leq\|\mu_1\|^2/2+\|\mu_2\|^2/2\leq\|\mu_1\|+\|\mu_2\|.
$$
