Based on the IMO

I'd like some feedback on this solution, as I'm not sure it's correct. The answer of $0$ is definitely correct, though.

First, I set values for $x$ and $y$ such that I had actual numerical values on the left side of the inequality. I used $x=k$ and $a=c-k$ , which allows me to rewrite things like this:

$f\left( c \right) \le \left[ c-k \right] f\left( k \right) +f\left( f\left( k \right) \right)$

I then defined $b=f\left( k \right)$ . Note that $b$ and $k$ are entangled because $k$ is part of the determination of $b$ . This will have to be worked around.

Either way, $c$ represents the domain of $f$ , and $b$ represents its range. I rewrote the modified inequality above as

$f\left( c \right) \le cb-kb+f\left( b \right)$

Which can be rewritten again as

$f\left( c \right) \le cb+f\left( 0 \right) -p$

Note that I randomly pulled a variable $p$ in. It is a positive (or zero-valued) real that merely serves the purpose of stating that $f\left( 0 \right)$ could be less than what I just replaced, but is what I just replaced at its maximum possible value.

Note that $f\left( 0 \right)-p$ will not change. It should be a set value for the function, so I can continue.

If the range, $b$ , of $f$ is NOT $0$ everywhere, then $c$ can cause the value of $f$ to have to continually decrease as $c$ decreases (or as $c$ increases for negative values). This would cause the range of $f$ to go to negative infinity and would allow $b$ to be a much smaller value that would eventually cause the inequality to no longer be true. Thus, the range of $f$ is $0$ , so $f\left( -1 \right) =0$ .