Consider below integral expression
$$\int_{0}^{\infty}g(y)[\int_{a}^{\infty}(1-e^{-(k+y)x})f(x)dx ]dy \ \ \ \ (1)$$
Where, we know: $$f(x)>0\ ,\ \ a\leq x \leq \infty$$ $$\ k>0$$ $$g(y)>0\ ,\ \ 0\leq y \leq \infty$$
I need to show $(1)$ has an upper-band as: $$\int_{0}^{\infty}g(y)[\int_{a}^{\infty}f(x)dx ]dy \ \ \ \ (2)$$
My Analysis:
I think since $f(x)$ is always positive,we can consider it as a density function. Also, since $e^{-(k+y)x}$ is strictly decreasing with respect to both $x$ and $y$, we can say average value of the $(1-e^{-(k+y)x})$ under density function of $f(x)$ is lower than the case setting the exponential term to zero. i.e., $e^{-(k+y)x}:=0$.
Question:
If I am right, how can I prove my physical analysis mathematically? ($(1) \leq (2)$):
Any hint, guidance, and comment would be appreciated.