Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $$f(x + y) \leq yf(x) + f(f(x))$$ for all real numbers $x$ and $y$. Prove that $f(x) = 0, \; \forall \; x \leq 0 .$
My attempts
I tried, $x=0, y=0, x=-y, y=-x$ and $x=y$, but I could not solve the problem.
$y=0$ then $f(x) \leq f(f(x))$
$x=0$ then $f(y) \leq yf(0)+f(f(0))$
$y=-x$, then $f(0)\leq -xf(x)+f(f(x))$
$x=-y$, then $f(0) \leq y(f(-y))+f(f(-y))$
$x=y$, then $f(2x) \leq xf(x)+f(f(x))$
Actually I tried more but I always failed. Can anyone solve this problem?