$f$ 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 $T &gt; 0$ 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?
Let $g(x)=\mathrm e^xf(x)$, then $g'(x)=\mathrm e^x(f'(x)+f(x))\geqslant0$ hence: $$(1)\ \textit{The function $g$ is nondecreasing.}$$
Since $f$ is continuous and periodic, $f$ is bounded, say $|f(x)|\leqslant C$ for every $x$, hence $|g(x)|\leqslant C\mathrm e^x\to0$ when $x\to-\infty$, that is: $$(2)\ \textit{The function $g$ has limit $0$ at $-\infty$.}$$
Properties (1) and (2) of $g$ imply together that $g\geqslant0$ everywhere, hence $f\geqslant0$ everywhere.



Examples: Consider $$f(x)=c\,\mathrm e^{w\cos(ux+v)},$$ for every  $c\geqslant0$, $u\ne0$, $|uw|\leqslant1$ and $v$, then $f$ has period $2\pi/|u|$ and, for every $x$, $$f'(x)+f(x)=(1-wu\sin(ux+v))\,f(x)\geqslant0.$$
For example, if $c=w=u=1$, $v=0$, one gets the function $f$:

$\qquad\qquad\qquad$

...And the function $f'+f$:

$\qquad\qquad\qquad$
$ 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