total variation distance of product of measures Let $f, \hat{f}, g,$ and $\hat{g}$ be continuous probability densities. Define probability densities $p \propto fg$ and $\hat{p} \propto \hat{f}\hat{g}$. Is it true that 
\begin{align*}
||p - \hat{p}||_{1} \le || f - \hat{f}|| _{1} + ||g - \hat{g}||_1
\end{align*}
 A: Here is a real counterexample, as verified by mathematica: so we take the same discrete two point space as below, and let $f=(u_1,u_2), \hat{f} = (v_1,v_2), g = (a_1,a_2), \hat{g}=(b_1,b_2)$, with the relation that $x_1 + x_2 = 1$ where $x \in \{u,v,a,b\}$. So your claim becomes in this special case
$$ |\frac{u_1 a_1}{u_1 p_1 + u_2 a_2} - \frac{v_1 b_1}{v_1 b_1 + v_2 b_2}| + | \frac{u_2 a_2}{u_1 a_1 + u_2 a_2} - \frac{v_2 b_2}{v_1 b_1 + v_2 b_2}|$$
$$\le |u_1 - v_1| + |u_2 - v_2| + |a_1 - b_1| + |a_2 - b_2|$$
Now using the relation that if $x + y = 1$ and $w + z = 1$, then $|x - w| + |y - z| = 2 |x -w|$, we can simplify the above inequality to
$$ |\frac{u_1 a_1}{u_1 a_1 + u_2 a_2} - \frac{v_1 b_1}{v_1 b_1 + v_2 b_2}| \le |u_1 - v_1| + |a_1 - b_1| $$.
What I then did is to subtract the square of the right hand side from the square of the left hand side, and call the resulting function $h(u,v,a,b)$, where $x := x_1$ for $x \in \{u,v,a,b\}$. Then I did some brute force search of maximum and found the following point:
$$u=v =0.456239, a = 1/5, b = 1/2$$ 
If you plug it in to $h$ you should get $0.000652285 > 0$. So that means the left hand side is not always $\le$ the right hand side. Here are the mathematica code so that you can help me verify:
q[u1_, u2_, a1_, a2_, v1_, v2_, b1_, 
   b2_] := (u1 a1/(u1 a1 + u2 a2) - 
      v1 b1/(v1 b1 + v2 b2))^2 - ( (u1 - v1)^2 + (a1 - b1)^2 + 
     2 Abs[(u1 - v1) (a1 - b1)]);
h[u_, v_, a_, b_] := q[u, 1 - u, a, 1 - a, v, 1 - v, b, 1 - b];
h[0.456239, 0.456239, 2/3, 2/5]
= 0.000652285
Edit: below was my earlier false counterexample. 
I believe it's false. Here is an argument. It suffices to consider discrete probability space, say the two point space $\Omega = \{ 0,1 \} $. Now let $f =(u_1,1-u_1)$ be the probability mass function, so that means $f(0) = u_1$ and $f(1) = 1-u_1$. and similarly $\hat{f} = (u_2,1-u_2)$. Also I will let $g = \hat{g} = (p, 1-p)$. Then $\|g - \hat{g}\|_1 = 0$, so your inequality amounts to 
$ |\frac{p u_1}{p u_1 + (1-p)(1-u_1)} - \frac{p u_2}{pu_2 + (1-p)(1-u_2)}| \le |u_1 - u_2|$.
Now take $u_1 = 2/3$ and  $u_2 = 1/3$, and let $ 1-p >> p$. Then you get a contradiction. 
