If $f'(x)>f(x)$ for all reals and $f(x_{0})=0$, then $f(x)>0$ for $x>x_{0}$. Someone had asked an interesting question but deleted it after one
hour. Since the question is interesting, I will re-post it and submit a
full solution.

Question: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable
function such that $f'(x)>f(x)$ for all $x\in\mathbb{R}$. Suppose
that there exists $x_{0}\in\mathbb{R}$ such that $f(x_{0})=0$, prove
that $f(x)>0$ for all $x>x_{0}$.
 A: Let $g(x)=e^{-x}f(x)\implies g(x_0)=0$. Since $g(x)$ is differentiable and $f'(x)>f(x),$
$$g'(x)=e^{-x}\left(f'(x)-f(x)\right)>0$$
Therefore, $g(x)$ is strictly increasing and $g(x)>0,\;\forall x>x_0$.
Also, $$f(x)=e^xg(x)>0$$
for all $x>x_0$.
A: Solution:
Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $g(x)=f(x+x_{0})$.
Clearly, $g$ is differentiable and for each $x\in\mathbb{R}$, $g'(x)=f'(x+x_{0})>f(x+x_{0})=g(x)$.
$g(0)=f(x_{0})=0$. We go to prove by contradiction that $g(x)>0$
for all $x>0$. Suppose that there exists $x>0$ such that $g(x)\leq0$.
Define $A=\{x\in(0,\infty)\mid g(x)\leq0\}$. Since $A$ is non-empty
and bounded below by $0$, $\xi:=\inf A$ exists. Firstly, we show
that $\xi=0$. Prove by contradiction. Suppose that $\xi>0$. Choose
a sequence $(x_{n})$ in $A$ such that $x_{n}\rightarrow\xi$. Since
$g$ is continuous and $g(x_{n})\leq0$, by letting $n\rightarrow\infty$,
we have that $g(\xi)\leq0$. By Mean-Value Theorem, where exists $c\in(0,\xi)$
such that $g(\xi)-g(0)=g'(c)(\xi-0)$. Hence, $g(c)<g'(c)=\frac{g(\xi)}{\xi}\leq0$
because $g(\xi)\leq0$ and $\xi>0$. Therefore $c\in A$. However,
$c<\xi$, contradicting to the fact that $\xi$ is a lower bound of
$A$. This shows that $\xi=0$.
Since $\xi\notin A$, we can choose a strictly decreasing sequence
$(x_{n})$ in $A$ such that $x_{n}\rightarrow\xi=0$. We have that
\begin{eqnarray*}
 &  & g'(0)\\
 & = & \lim_{x\rightarrow0}\frac{g(x)-g(0)}{x-0}\\
 & = & \lim_{n\rightarrow\infty}\frac{g(x_{n})-g(0)}{x_{n}-0}\\
 & = & \lim_{n\rightarrow\infty}\frac{g(x_{n})}{x_{n}}\\
 & \leq & 0
\end{eqnarray*}
because $g(x_{n})\leq0$ while $x_{n}>0$. On the other hand, $g'(0)>g(0)=0$,
which is a contradiction.
