TITLE: Show that $\frac{f(x)}{x}$ is a decreasing function implies that $f(x)$ is subadditive
QUESTION [7 upvotes]: 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.

REPLY [10 votes]: 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
$$\frac{f(y)}{y} \geq \frac{f(x+y)}{x+y} \; \implies f(y) \geq \frac{yf(x+y)}{x+y} \tag{2}\label{eq2}$$
Adding \eqref{eq1} and \eqref{eq2} gives
$$f(x) + f(y) \geq \frac{xf(x+y)}{x+y} + \frac{yf(x+y)}{x+y} = \frac{(x+y)f(x+y)}{x+y} = f(x+y) \tag{3}\label{eq3}$$
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.