Consider the functional inequalities:
$$ (1)\; f(x)\geq f(y)(1+x-y) \; ; \; x,y\in \mathbb{R}, $$ $$ (2)\; \frac{f(x)}{f(y)}\geq 1+x-y \; ; \; x,y\in \mathbb{R}. $$
It is obvious that all functions of the form $f(t)=ce^t$ with $c\geq 0$ (resp. $c\neq 0$) satisfy $(1)$ (resp. $(2)$).
Are there any other solutions for them?
Thnaks in advance
We deal with (2). For any given $t,\delta\in\mathbb R$, we have $$\frac{f(t+n\delta)}{f(t)}=\prod_{i=0}^{n-1}\frac{f(t+(i+1)\delta)}{f(t+i\delta)}\geq (1+\delta)^n,$$ so $$\frac{f(x)}{f(y)}\geq \lim_{n\to\infty}\left(1+\frac{x-y}{n}\right)^n=e^{x-y}.$$ Furthermore, as $e^v\geq 1+v$ for all real $v$, this is sufficient for (2). However, $$\frac{f(y)}{f(x)}\geq e^{y-x};$$ these two inequalities multiply to $1\geq 1$ (and we can multiply them since everything is positive), and so equality must hold; as a result, $f(x)=ce^x$ for some constant $c$.
For (1), the process is very similar, and the answer is only slightly different, in essentially the way that the solutions you found differ (values of $0$ are acceptable; negative values are not).