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*}

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 $f =(u_1,1u_1)$ be the probability mass function, so that means $f(0) = u_1$ and $f(1) = 1u_1$. and similarly $\hat{f} = (u_2,1u_2)$. Also I will let $g = \hat{g} = (p, 1p)$. Then $\g  \hat{g}\_1 = 0$, so your inequality amounts to $ \frac{p u_1}{p u_1 + (1p)(1u_1)}  \frac{p u_2}{pu_2 + (1p)(1u_2)} \le u_1  u_2$. Now take $u_1 = 2/3$ and $u_2 = 1/3$, and let $ 1p >> p$. Then you get a contradiction. 

