I am trying to show that if $f:(0,+\infty)\rightarrow\mathbb R$ is a $C^1$ function such that $$f'(x)<\frac{f(x)}{x}\quad \forall x\in (0,+\infty) \tag{$\star$}$$ then $$f(x+y)<f(x)+f(y)$$
First of all, I showed that
$x\mapsto \frac{f(x)}{x}$ is a decreasing function.
Its derivative is $$\frac{d}{dx}\left (\frac{f(x)}{x} \right )=\frac{xf'(x)-f(x)}{x^2}=\frac{1}{x}\left (f'(x)-\frac{f(x)}{x} \right )$$ Since $f$'s domain is $(0,+\infty)$, $x\mapsto \frac{f(x)}{x}$ is decreasing because of $(\star)$.
If $x,y\in(0,+\infty)$ are two positive real numbers such that $x\leq y$, then $f'(y)<\frac{f(x)}{x}$.
Because of $(\star)$, $$f'(y)<\frac{f(y)}{y}$$ The result follows from the fact that $x\mapsto \frac{f(x)}{x}$ is decreasing.
I proved the result through the following integral estimate. Let $x\leq y$. We can write $$f(x+y)-f(y)=\int_{y}^{x+y}f'(t)dt$$ Every $t$ in the interval $(y~,~x+y)$ is greater than $y$, so we know that $$f'(t)<\frac{f(y)}{y}$$ from the second point. So $$\int_{y}^{x+y}f'(t)dt<\frac{f(y)}{y} \cdot x$$ Since $x\leq y$, we know from the first point that $$\frac{f(y)}{y} \cdot x<\frac{f(x)}{x}\cdot x=f(x)$$ so we have proven that $$f(x+y)-f(y)<f(x)$$ Is the claim true? Are there any flaws in this proof?
Since $$\frac{f(x)}{x} $$ is decreasing we have $$\frac{xf(x+y)}{x+y} < f(x) \wedge \frac{yf(x+y)}{x+y} < f(y)$$ adding both these inequalities we get $$f(x) +f(y) > \frac{xf(x+y)}{x+y} +\frac{yf(x+y)}{x+y}= f(x+y) $$