$f$ 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?
 A: 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 $f'+f$:
$\qquad\qquad\qquad$
A: We don't even need that $f'(x)$ is continuous! It is enough to show that $f(x) \ge 0$ on $[0,T]$, because of periodicity. Since $[0,T]$ is compact, $f$ has a minimum on $[0,T]$. Let $x_0 \in [0,T]$ be such that $f(x_0)$ is minimal. If $x_0 =0$ or $x_0 =T$, then $f(x_0)$ is also minimal in $[-T,2T]$, because of periodicity.
A necessary condition for a minimum is that $f'(x_0) =0$. (And it is also valid in the case $x_0 =0$ or $x_0 =T$, because of the above remark.) Thus we get $f(x_0) = f(x_0)+f'(x_0) \ge 0$. This shows that $f(x) \ge 0$ everywhere.
A: Since $f$ is periodic, it has maximum and minimum. Choose the period $[a,b]$, such that $f(a)=f(b)=M$, where $M$ is the maximum.
It can be seen that, there is a point $f(c)=m$, where $m$ is the minimum. c is golbal minimum, and thus local minimum, so:
$$f'(c)=0$$
from your condition.
$$f+f'\ge 0$$
we see 
$$f(c)\ge 0$$. Since $f(c)$ is the minimum, we are done!
A: Consider such $f(x)$ to be non-constant. There must be a global minimum within each period, say at $x=x_0$. Since $f(x)$ is differentiable and continuous, $f'(x_0) = 0$. Hence 
$$f(x) \ge f(x_0) = f(x_0)+f'(x_0) \ge 0$$ 
A: Say a period is $[a,a+T]$. Since $f$ is continuous, it attains a minimum on that interval, say at $c$. (If you have $c = a$, then change the period to $[a - T/2,a + T/2]$ so that $c$ becomes an interior point.)
We must have $f'(c) = 0$ there, so $f(c) \geq 0$. So the minimum value of $f$ is nonnegative.
A: suppose $f(x)$ is negative for all values of $x$. Periodicity tells us that there must be solutions to $f'(x)=0$ and for such $x$ we would then have $f(x)+f'(x)=f(x)<0$.
Suppose that $f(x)$ is negative for some values of $x$ but not all.  Let $a,b$ be zeroes of $f(x)$ such that $a<x<b\implies f(x)<0$. Specifically, let $x_0$ be some value for which $f(x_0)<0$ and let $a$ be the greatest upper bound of $\{x\,:\,x<x_0\,\;\&\,\;f(x)≥0\}$ and $b$ defined similarly on the other side. Since $f(a)=0=f(b)$ there is some $c$ with $a<c<b$ and $f'(c)=0$.  But for that value we must have $$f(c)+f'(c)=f(c)<0$$ contrary to assumption.
A: $ 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
