I found that $f$ it is identical $0$, it is true? If yes, I will present my solution.

Let $P(x,y)$ be $f(f(x) - y) \le xf(x) + f(y)$. 1. $xf(x) \ge 0 \,\forall x \in R$ ======================== Proof: Let $f(x) - y = t$. Now from $P(x,y)$ we have: $f(t)\le xf(x) + f(f(x)-t)\le xf(x) + xf(x) + f(t)$ and then $f(t)\le2xf(x)+f(t)$. Now $xf(x)\ge 0$. Note that $ x>0 \implies f(x) \ge 0$ and $ x<0 \implies f(x) \le 0$ 2. $f$ has at least one nullpoint ======================== Proof: From $P(0,y)$ we have $f(f(0)-y) \le f(y)$. If $f(0)=0$ we are done so suppose $f(0) \neq 0$ Case 1: $f(0)>0$. For any $y<0 \implies f(0)-y >0$. Now $f(f(0)-y) \le f(y) \implies 0 \le f(f(0)-y) \le f(y) \le 0 \implies f(f(0)-y) = 0$ Case 2: $f(0)<0$. For any $y<f(0)<0 \implies f(0)-y>0$. Now $f(f(0)-y) \le f(y) \implies 0 \le f(f(0)-y) \le f(y) \le 0 \implies f(f(0)-y) = 0$ 3. $f(y) = 0 \,\forall y\neq0$ ======================== Proof: Let $a$ be one nullpoint of $f$. Then from $P(a,y)$ we have $f(-y) \le f(y)$. If $y<0$ we have $0 \le f(-y) \le f(y) \le 0 \implies f(-y) = f(y) =0$ 4. $f(0) = 0$ ======================== Proof: Suppose that $f(0) \neq0 $ Case 1: $f(0) = t > 0$. From $P(0,t)$ we have $f(0) \le 0$, contradiction. Case 2: $f(0) = t < 0$. From $P(0,0)$ we have $f(0) \ge 0$, contradiction. Now we have $f(x) = 0 \,\forall x \in R$ and we can verify it is a solution.

parmenides51 wrote: Determine all functions $f : R \to R$ satisfying $f(f(x) - y) \le xf(x) + f(y)$ for all real numbers $x, y$. https://artofproblemsolving.com/community/c6h1447208p8275534