Consider a function $f$ that is strictly concave and differentiable on $\mathbb{R}_{+}$ such that $f(0)=0$. I would like to show that $\frac{f(x)}{x}$ is strictly decreasing in $\mathbb{R}_{+}^{*}$
Here is my attempt : since $f(x)$ is strictly concave we know that for all $x,y\in\mathbb{R}_{+}$ we have
$$ x\neq y \implies f(y) - f(x) > f'(y)(y-x) $$
and taking $x=0$ we get $\frac{f(y)}{y}> f'(y)$ for all $y\in\mathbb{R}_{+}^{*}$ but I don't see how to continue (if it is the right way) ? I would like to use the fact that $f'(y)$ is strictly decreasing but as far as I know I can just write that
$$ x>y \implies \frac{f(y)}{y}> f'(y) > f'(x) $$
Thank you a lot !
