This question was originally asked in stats.stackexchange (the link is given below). I'm reposting it here as I did not get any answer to the said post. https://stats.stackexchange.com/questions/423833/proof-of-likelihood-ratio-ordering-implying-hazard-rate-ordering
I'm studying stochastic orders and am stuck in the proof of likelihood ratio ordering implying hazard rate ordering.
Definitions: Let $X$ and $Y$ be two random variables with respective densities $f(t)$ and $g(t)$, and respective survival functions $\bar{F}(t):=\int_t^{\infty} f(u)\,du$ and $\bar{G}(t):=\int_t^{\infty} g(u)\,du$. Then we say that: $(i)$ $X \leq_{HR} Y$ if $\bar{G}\left(t\right)/\bar{F}\left(t\right)$ is nondecreasing in $t$ and $(ii)$ $X \leq_{LR} Y$ if $g\left(t\right)/f\left(t\right)$ is nondecreasing in $t$.
The Problem: To show that $X \leq_{LR} Y \implies X \leq_{HR} Y$.
My Progress: We assume that $X \leq_{LR} Y$. Then by definition, for every $t<t'$, we have $$\frac{g(t)}{f(t)} \leq \frac{g(t')}{f(t')},$$
i.e. $f(t)g(t') \geq f(t')g(t)$. Let $x \leq y$. Integrating over $(x, y) \times (y, \infty)$, we have $$\int_x^y \int_y^{\infty} f(t)\,g(t')\,dt'\,dt \geq \int_x^y \int_y^{\infty} f(t')\,g(t)\,dt'\,dt \tag{1}$$
I want to achieve: $$\int_x^{\infty} \int_y^{\infty} f(t)\,g(t')\,dt'\,dt \geq \int_x^{\infty} \int_y^{\infty} f(t')\,g(t)\,dt'\,dt, \tag{2}$$
so that we can separate out the integrals and write $$\int_x^{\infty} f(t)\,dt \int_y^{\infty} g(t')\,dt' \geq \int_y^{\infty} f(t')\,dt' \int_x^{\infty} g(t)\,dt,$$
i.e. $\bar{F}(x)\bar{G}(y) \geq \bar{F}(y)\bar{G}(x)$ for $x \leq y$. Since, the choice of $x$ and $y$ are arbitrary, $\bar{G}/\bar{F}$ is increasing, hence $X \leq_{HR} Y$.
Question: How do I come to $(2)$ from $(1)?$ Thank you.
P.S. If there is any other way to prove this result, please do share the source where I can find it.