# Prove $f'(x) \geq x f(x)$ $\forall x \in \mathbb{R}$ $\implies$ $\exists k$ s.t. $ke^{x} \leq f(x)$ $\forall x \in \mathbb{R}$.

Question:

Let $f$ be a differentiable function. Prove $f'(x) \geq x f(x)$ $\forall x \in \mathbb{R}$ $\implies$ $\exists k$ s.t. $ke^{x} \leq f(x)$ $\forall x \in \mathbb{R}$.

I've made some attempts. I know that if we take the derivative of $f(x) = e^{kx}$, we get $f'(x) = ke^{kx}$, which is of a similar form to the above. Although, I'm not sure if this really helps me much. Any suggestions?

I think there is a problem with your problem. Try $f(x)=-e^{(\frac{x^2}{2}-x)}$