Proof of direct proportionality between hazard rate function and probability density function I'm reading up on reliability and I came across this question:

Show that if the hazard function is decreasing, the PDF, $f(t)$, is also a decreasing function and its mode must therefore occur at t = 0

I know that:
$h(t)\ = \frac{f(t)}{S(t)}$
where $h(t)$ is the hazard rate, $f(t)$ is the pdf and $S(t)$ is the reliability.
From the formula, we can infer that $f(t)$ increases with $h(t)$ so it's directly proportional and $s(t)$ decreases so it's indirectly proportional.
I also know that the mode of a probability density function is the point at which the maximum value occurs so I figure we're meant to derive some sort of formula and then set t=0 to get the maximum value, thereby proving that it is the mode.
What I'd like to know is how to derive a formula that proves that PDF decreases with the hazard function.
 A: It's tempting to show the derivative of $f$ must be non-positive: but there's no assurance $f$ is differentiable.  Let's therefore attempt a direct comparison inspired by the finite difference: that is, to say that a function $f$ is decreasing literally means for all $t \ge 0$ and $\epsilon \gt 0,$
$$f(t+\epsilon) - f(t) \lt 0.$$
That's what we need to show.  There's nothing available to try except to plug in the definitions.  In the following, $S(t) = \int_t^\infty f(x)dx$ is the survival  function, which we must assume to be nonzero.  Let's write down the finite difference for $h,$ which we assume is negative:
$$\eqalign{
0 \gt h(t+\epsilon) - h(t)  &= \frac{f(t+\epsilon)}{S(t+\epsilon)} - \frac{f(t)}{S(t)}\\
&=\frac{f(t+\epsilon)S(t) - f(t)S(t+\epsilon)}{S(t+\epsilon)S(t)}.
}$$
Clearly both $S(t)$ and $S(t+\epsilon)$ are positive, so we may ignore them when considering just the sign of the fraction.  Let's focus on its numerator:
$$0 \gt f(t+\epsilon)S(t) - f(t)S(t+\epsilon) = \color{red}{(f(t+\epsilon)-f(t))S(t)} - \color{blue}{f(t)(S(t+\epsilon)-S(t))}.$$
This can be more simply written
$$\color{blue}{f(t)(S(t+\epsilon)-S(t))} \gt \color{red}{(f(t+\epsilon)-f(t))S(t)}.$$
Since $S(t+\epsilon) - S(t) = -\int_t^{t+\epsilon}f(x)dx \le 0$ and $f(t)\ge 0,$ the left hand side cannot be positive.  The factor of $S(t)$ on the right hand side is positive (because it is nonzero).  Therefore, the term it multiplies must be negative:
$$0 \gt f(t+\epsilon) - f(t),$$
QED.
