$f$ has a second derivative $f'' < 0$ $\implies$ $f$ has a decreasing first derivative $\implies$ $\frac{f(x)}{x}$ is decreasing for $x > 0$. I am currently working on the following problem and had a few questions regarding my work:

if $f$ has a second derivative $f'' < 0$ $\implies$ $f$ has a decreasing first derivative. Show this implies that   $\frac{f(x)}{x}$ is decreasing for $x > 0$.

My Work Thus Far:
Let $g(x) = \frac{f(x)}{x}$, for $x >0$. According to the question we have that $f''<0$ then $f$ has a decreasing first derivative, meaning $f'$ is decreasing. Now, taking the derivative of $g$ yields: $g'(x) = \frac{xf'(x) - f(x)}{x^{2}}$ $=$ $\frac{f'(x)}{x} - \frac{f(x)}{x^{2}}$, where $x >0$.
I know that $ - \frac{f(x)}{x^{2}}$ is decreasing, but how do I know for sure that $\frac{f'(x)}{x}$ is decreasing so I can deduce that $g'(x) <0$, $\forall x >0$ and that $g(x)$ is a decreasing function?

Remark: I think what has really confused me is why can we say $\frac{f'(x)}{x}$ is decreasing? Is this simply because $f'(x)$ is decreaing? If this is the case, why?  $f'(x)$ and $\frac{f'(x)}{x}$ are two different functions.
For the sake of context, this question was pulled from the problem: let $f: [0, \infty) \to [0, \infty)$ be increasing and satisfy $f(0) = 0$ and $f(x) > 0$ $\forall x >0$. If $f$ also satisfies $f(x+y) \leq f(x) + f(y)$ $\forall x,y \geq 0$, then $f \circ d$ is a metric whenever $d$ is metric. Show each of the following conditions is sufficient to ensure that $f(x+y) \leq f(x) +f(y)$ $\forall x,y \geq 0$:
$a)$ $f$ has a second derivative satisfying $f'' \leq 0$;
$b)$ $f$ has a decreasing first derivative.
$c)$ $\frac{f(x)}{x}$ is decreasing for $x > 0$.
To prove this claim, I figured out that it would be easier to show that $a)$ $\implies$ $b)$ $\implies$ $c)$; hence where my question arose.
 A: Geometric intuition: $f(x)/x$ is the slope of the line connecting $(x,f(x))$ to the origin.
This leads us to a simple counterexample for the original claim: with $f(x) = -(x-2)^2$ we have $f(1)/1 = -1$ and $f(2)/2 = 0$.
However, the claim is true with the added condition $f(0)=0$.
The concavity of $f$ (given by $f'' < 0$) then implies that this slope decreases as $x$ increases.

Let $0<x<y$. If we show $$\frac{f(x)}{x} \ge \frac{f(y)-f(x)}{y-x},\tag{$*$}$$ then we have $$f(y) = f(x) + \frac{f(y)-f(x)}{y-x} (y-x) \le f(x) + \frac{f(x)}{x} (y-x) = y \frac{f(x)}{x}$$ which is what we want.
To prove ($*$), note that the mean value theorem implies $\frac{f(x)}{x} = \frac{f(x)-f(0)}{x-0} = f'(a)$ for some $0 \le a \le x$ and $\frac{f(y)-f(x)}{y-x} = f'(b)$ for some $x \le b \le y$. Using the fact that $f'$ is decreasing, we have $f'(a) \ge f'(x) \ge f'(b)$ which proves ($*$).
A: As mentioned above this is true for $f(0)\ge 0$ Assuming this condition:
now continuing from your attempt $g'(x)=\frac{xf'(x)-f(x)}{x^2}$
let us consider $h(x)=xf'(x)-f(x)$
$h'(x)=xf''(x)<0$ or $h(x)$ is decreasing  for all positive x
Also $h(0)<0$ using $f(0)\ge 0$ or $h(x)<0$ for all positive x.
this implies $g'(x)=\frac{h(x)}{x^2}<0$ for all positive x
