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\*}



total variation distance of product of measures
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. 


