Prove that $$\mbox{Var}_g[\frac{f(x,y)}{g(x,y)}]\geq \mbox{Var}_g[\frac{f_1(x)}{g_1(x)}]$$ where $$f_1(x) = \int f(x,y) \mathrm{d}y$$ and $f$ and $g$ are distributions.
Here's my attempt:
the LHS is
$$\int \int \frac{f^2}{g^2}g\mathrm{d}x\mathrm{d}y - (\int \int \frac{f}{g}g\mathrm{d}x\mathrm{d}y)^2$$ $$=\int \int \frac{f^2}{g}\mathrm{d}x\mathrm{d}y - (\int \int f\mathrm{d}x\mathrm{d}y)^2$$ $$=\int \int \frac{f^2}{g}\mathrm{d}x\mathrm{d}y - 1.$$
The RHS is
$$\int \int \frac{f_1^2}{g_1^2}g\mathrm{d}x\mathrm{d}y - (\int \int \frac{f_1}{g_1}g\mathrm{d}x\mathrm{d}y)^2$$
$$=\int \frac{f_1^2}{g_1^2}g_1\mathrm{d}x - (\int \frac{f_1}{g_1}g_1\mathrm{d}x)^2$$ $$=\int \frac{f_1^2}{g_1^2}g_1\mathrm{d}x - (\int f_1\mathrm{d}x)^2$$
$$=\int \frac{f_1^2}{g_1}\mathrm{d}x - 1$$
Is there any reason why
$$\int \int \frac{f^2}{g}\mathrm{d}x\mathrm{d}y$$
should have to be greater or equal to
$$\int \frac{f_1^2}{g_1}\mathrm{d}x$$?
From Cauchy Schwarz inequality,
$$(\int\dfrac{f^2(x,y)}{g(x,y)}dy)(\int g(x,y)dy)\geq (\int f(x,y)dy)^2$$
This yields
$$\int \dfrac{f^2(x,y)}{g(x,y)}dy\geq\dfrac{f_1^2(x)}{g_1(x)}$$
Now integrate both sides w.r.t. $x$ to get your inequality.