The.Math.Terminator wrote:

What if $f(x)$ is negative for $x>1$?

@OP are you sure you don't have any typos?

here it is

I claim all $f$ are of the following form: $f(x)\ge 0$ for $x\le 0$ and $f(x)=0$ for $x>0$. Denote the given assertion by $P(x,y)$. Then $P(x,0)$ yields $f(x)(1-x)\ge 0$. From here, if $x>1$ then $f(x)\le 0$ and if $x<1$ then $f(x)\ge 0$. Moreover, $P(x,1)$ implies $f(x+1)\ge xf(x)+f(1)$. Taking $x\in(0,1)$ we get $0\ge f(x+1)-xf(x)=f(1)$. Thus $f(x)\le 0$ for $x\ge 1$. Now, $P(x,x)$ implies $f(2x)\ge 2xf(x)$. In particular, for any $1/2\le x\le 1$, we get $0\ge f(2x)\ge 2xf(x)$, thus $f(x)\le 0$. Hence $f(x)=0$ for $1/2\le x\le 1$ and $1\le x\le 2$. Iterating this identity, we find $f(x)=0$ for every $x>0$ and $f(x)\ge 0$ for $x\le 0$ (as above). We now show this function indeed obeys the conditions. If $x,y\ge 0$, then $f(x+y)=0$ whereas $xf(x)+yf(y)=0$, so we trivially have the inequality. Likewise if $x,y\le 0$ then $f(x+y)\ge 0$ whereas $xf(x),yf(y)\le 0$, so once again we are done. Finally, let $x\ge 0$ and $y\le 0$. Then $f(x+y)\ge 0$. Moreover, $xf(x)=0$ whereas $yf(y) \le 0$, so we are done once again.