parmenides51 wrote: Find all functions $ f: \mathbb {R} \rightarrow \mathbb {R} $ such that $ f (x + y) \le f (xy) $ for every pair of real $ x $, $ y$. Equation is Assertion $P(u,v)$ : $f(u)\le f(v)$ $\forall u,v$ such that $u^2\ge 4v$ Let $x,y\in\mathbb R$ and $z<0$ small enough to have $z^2\ge 4x$ and $z^2\ge 4y$ $z^2\ge 4x$ and so $P(z,x)$ $\implies$ $f(z)\le f(x)$ $z\le 0$ $\implies$ $x^2\ge 4z$ and so $P(x,z)$ $\implies$ $f(x)\le f(z)$ And so $f(z)=f(x)$ $z^2\ge 4y$ and so $P(z,y)$ $\implies$ $f(z)\le f(y)$ $z\le 0$ $\implies$ $y^2\ge 4z$ and so $P(y,z)$ $\implies$ $f(y)\le f(z)$ And so $f(z)=f(y)$ And so $f(x)=f(y)$ $\forall x,y$ and $\boxed{f(x)=c\quad\text{constant}}$, which indeed fits, whatever is $c\in\mathbb R$

$P(x,0)\Rightarrow f(x)\le f(0)$ $P(x,-x)\Rightarrow f(-x^2)\ge f(0)\Rightarrow f(x)=f(0)\forall x\le0$ Let $x>0$: $P(-x,-x)\Rightarrow f(x^2)\ge f(-2x)=f(0)$ Thus $\boxed{f(x)=c}$ for all $x$.