# Show that $\frac{f(x)}{x}$ is a decreasing function implies that $f(x)$ is subadditive

I am studying Carother's Real Analysis for my qualifying exams. In the book I am to prove that if $$f : [0, \infty) \rightarrow [0, \infty)$$ is an increasing function, $$f(0) = 0$$, and $$f(x) > 0$$ for all $$x > 0$$, then $$\frac{f(x)}{x}$$ being decreasing for $$x > 0$$ implies that $$f$$ is subadditive, or that $$f(x + y) \leq f(x) + f(y)$$.

So far I have tried:

$$\frac{f(x+y)}{x+y} \leq \frac{f(y)}{y}$$ and $$\frac{f(x+y)}{x+y} \leq \frac{f(x)}{x}$$ implies that $$2\frac{f(x+y)}{x+y} \leq \frac{f(x)}{x} +\frac{f(y)}{y}$$, so $$\frac{f(x+y)}{x+y} \leq 2\frac{f(x+y)}{x+y} \leq \frac{f(x)}{x} +\frac{f(y)}{y} \leq f(x) + f(y)$$

I think that I am close but I can't get rid of the $$x + y$$ in the denominator. Any help would be appreciated.

For $$x \gt 0$$ and $$y \gt 0$$, you get

$$\frac{f(x)}{x} \geq \frac{f(x+y)}{x+y} \; \implies f(x) \geq \frac{xf(x+y)}{x+y} \tag{1}\label{eq1}$$

and

$$\frac{f(y)}{y} \geq \frac{f(x+y)}{x+y} \; \implies f(y) \geq \frac{yf(x+y)}{x+y} \tag{2}\label{eq2}$$

$$f(x) + f(y) \geq \frac{xf(x+y)}{x+y} + \frac{yf(x+y)}{x+y} = \frac{(x+y)f(x+y)}{x+y} = f(x+y) \tag{3}\label{eq3}$$
For $$x$$ and/or $$y$$ being $$0$$, since $$f(0) = 0$$, then \eqref{eq3} still holds. Overall, this is what you were asked to prove, i.e., that $$f$$ is subadditive.
Note this didn't use that $$f$$ is an increasing function. Just having $$f(0) = 0$$ and $$\frac{f(x)}{x}$$ being a decreasing function is sufficient.