$f$ is T periodic and $f(x) + f'(x) \ge 0 \Rightarrow f(x) \ge 0$ 
Let $f: \Bbb R \to \Bbb R$ be a function such that $f'(x)$ exists and is continuous over $\Bbb R$. Moreover, let there be a $T > 0$ such that $f(x + T) = f(x)$ for all $x \in \Bbb R$ and let $f(x) + f'(x)\ge 0$ for all $x \in \Bbb R$.
Show that $f(x) \ge 0$ for all $x \in \Bbb R$.

My attempt:
$f(x) \ge 0 \iff f(x) \ge f'(x) - f'(x) \iff f(x) + f'(x) \ge f'(x)$.
Thus, it is enoguh to show that $0 \ge f'(x)$.
$\iff 0 \ge \lim_{h\to0}\frac{f(x + h) - f(x)}{h}$
I do not know how to proceed from here. I know that $f'$ also has a periodicity of T but I do not know how to use that here.
Am I on the right track? How can I use the periodicity of $f$ to solve the problem?

$ f \ne 0$
$f(x)=0 \Rightarrow f'(x) \ge 0$, hence $f$ can cross the X-axis at most once. periodicity means that it cannot cross  at all.
if $f$ is non-positive then $f' \ge 0$ so $f$ is monotone increasing. again this contradicts periodicity
