Find all functions $f$ over reals such that $f(xy) \geq f(x+y)$ for all $x,y \in \mathbb{R}$.
We have that $f(x) \geq f(x+1)$ and $f(0) \geq f(1)$. I am wondering how to use these conditions to solve the problem. It seems like using a telescoping sum might be useful with the $f(x) \geq f(x+1)$ condition.
We have $f(0) \geq f(y)$ for all $y$, setting $x=0$.
Now $f(0) \geq f(-x^2) \geq f(x-x) = f(0)$, so equality holds throughout and $f$ must be constant on $x \leq 0$.
And if $x>0$, $$f(0) \geq f(x) = f((-\sqrt{x})^2) \geq f(-2 \sqrt{x}) = f(0)$$ so equality holds throughout and $f$ is constant on $\mathbb{R}$.
It is easy to see that any constant will do, so the answer is that "$f(x) = r$ for all $x$" is a solution for each $r \in \mathbb{R}$, and it is the only possible set of solutions.