Another functional inequality

Question

Is there some general solution to the functional inequality:

$$ f(xy) \leq y f(x) + x f(y)$$

Where $x,y\in[0,1]$?

I can find many particular solutions but I just wonder if there is a more general description of functions f satisfying such functional inequality. I have the following conditions

1) $f : [0,1]\rightarrow \mathbb R$

2) $f$ is non-negative on $[0,1]$

3) $f(0) = 0$ and $f(1)$ is positive and finite, let say normalised to $1$

4) $f$ is one time continuously differentiable on $(0,1)$

Thanks for any suggestions

Answer 1

Robert gave me an idea with his comment "...nondecreasing nonnegative functions....".
In fact if I divide my original inequality by the product x*y my inequality become:
f(xy) <= f(x) + f(y) and then a general answer to my question can be:

Take a non decreasing nonnegative function g(x). Then the function f(x) = xg(x) satisfies the inequality f(xy) <= xf(y) + yf(x)

Thanks for your help Gianfranco