How to proof the following Gronwall type inequality? Suppose that $g,k:  [0,a] \to \mathbb R$ are continuous, $a >0 $, $\,k(t) \ge 0$,$\ c(t) \in C^1([0,a])$, $\, \dot c(t) \ge 0  $ (i.e. $c(t)$ is non decreasing) and  $g(t)$ satisfies $$g(t) \le c(t) + \int^{t}_{0} k(s) g(s)ds$$ for all $0  \le t \le a$. 
I want to show that for all $t \in [0,a]$, $$ g(t) \le c(t)e^{\int^{t}_{0} k(s) ds} $$ 
I have noticed that there is a proof for a more general case in wikipedia. However, I do not quite understand the proof and since the above is a less general case, I would guess that there is a simpler way to prove it. 
Let $G = c(t) + \int^{t}_{0} k(s) g(s)ds $. Then $G \in C^1([0,a])$. Taking the derivative, $$\dot G = \dot c + k(t)g(t) \\  \dot c =  \dot G - k(t)g(t)  \ge \dot G - k(t)G(t)$$. 
I am stuck here and not sure if I am going in the right direction. I suppose the goal is to reach $$ \frac{d}{dt} (G(t)e^{-\int^{t}_{0} k(s)ds}) \le\frac{d}{dt} c(t) $$ and integrate both sides. Any hints would be appreciated. 

You are almost there. Observe that 
\begin{align*}
\dot c =  \dot G - k(t)g(t) &\geq \dot G - k(t)G(t)\\
&= \exp\left(\int_0^t k(s)ds\right)\frac{d}{dt}\left\{G(t)\exp\left(-\int_0^t k(s)ds\right)\right\}\\
&\geq\frac{d}{dt}\left\{G(t)\exp\left(-\int_0^t k(s)ds\right)\right\}.
\end{align*}
