Let $f$ be a function from the set of real numbers to itself that satisfies $f(x + y) ≤ yf(x) + f(f(x))$ for all real numbers $x$ and $y$. Prove that $f(x) = 0$ for all $x ≤ 0$.
I tried to show that $f(x)\ge 0$ and $f(x)\le 0$ for all $x\le 0$ but i don't know how to derive from the inequality.