Romanian Mathematical Olympiad

The inequality $e^y\ge 1+y$, holds for all $y\in \mathbb{R}$. Assume the function $g$ is monotonously increasing on $\mathbb{R}$. Then, based on the inequality above, we get $$f(x+y)=e^{x+y}g(x+y)\ge e^x(1+y)g(x)=(1+y)f(x),$$ for all $x\in \mathbb{R}$ and all $y\ge 0$.

Conversely, assume $f(x+y)\ge (1+y)f(x)$, for all $x\in \mathbb{R}$ and all $y\ge 0$. One can check by induction $$f(x+nt)\ge (1+t{)}^{n}f(x),$$ for all $x\in \mathbb{R}$, $t\ge 0$ and $n\in \mathbb{N}$. Let $x,z\in \mathbb{R}$, with $x<z$. Denote $y=z-x$. For $n\in {\mathbb{N}}^{\ast }$ we have $$g(z)=e^{-z}f(z)=e^{-x-y}f\left(x+n\frac{y}{n}\right)\ge e^{-x-y}{\left(1+\frac{y}{n}\right)}^{n}f(x)=\frac{{\left(1+\frac{y}{n}\right)}^n}{e^y}g(x).$$ It follows $$g(z)\ge \underset{n\to \infty }{\lim }\frac{{\left(1+\frac{y}{n}\right)}^n}{e^y}g(x)=g(x),$$ therefore $g$ is monotonously increasing on $\mathbb{R}$.