$d_{TV}$ VS correlation coefficient. Consider two RV $X,Y$. If $d_{TV}(X,Y)=0$ you may couple them in such a way that $$\rho_{XY}=\frac {\operatorname{COV}(X,Y) }{\sigma_X\sigma_Y}=1.$$
So, is there any formula to bound $d_{TV}$ in terms of $\rho$?
 A: Here I do not provide a bound but a relation that can yield bounds.
Let the total variation distance between two probability measures defined on a space $(\Omega,\mathcal F)$ be 
$$
d_{TV}(P,Q)=2\sup_{A\in\mathcal F}|P(A)-Q(A)|.
$$
The coupling inequality shows that for all coupling $(X,Y)$ such that:
$$
d_{TV}(P,Q)\leq 2P(X\neq Y). 
$$
The maximal coupling theorem shows that the equality can be achieved by a coupling. So if $P(X\neq Y)$ can be related to $\rho$, then the problem is solved. Without loss of generality assume $X$ and $Y$ are zero-mean unit variance random variables:
$$
\rho=E(XY)=P(X\neq Y)E(XY|X\neq Y)+P(X=Y)E(XY|X=Y).
$$ 
But we have:
$$
P(X=Y)E(XY|X=Y)+P(X\neq Y)E(X^2|X\neq Y)=E(X^2)=1
$$
and
$$
P(X=Y)E(XY|X=Y)+P(X\neq Y)E(Y^2|X\neq Y)=E(Y^2)=1.
$$
Combining three above equations, we get:
$$
2(1-\rho)=P(X\neq Y)E[(X-Y)^2|X\neq Y].
$$
which means that for maximal coupling of $(X,Y)$ we have:
$$
4(1-\rho)=d_{TV}(P,Q)E[(X-Y)^2|X\neq Y].
$$
Now if $X$ and $Y$ are bounded random variables with $|X|\leq M$ and $|Y|\leq M$, then:

$$
(1-\rho)\leq M^2 d_{TV}(P,Q).
$$

There might be other ways of bounding $\rho$ however this is not clear at the moment to me. 
