Satisfying the inequality of a bounded derivative Hi I am having an issue of proving this inequality.  
The problem:
Suppose: $G(x) = |x|^2 + 1$
Show: $\left|\frac{d}{dx}G(x)\right| \leq cG$ (A bounded derivative)
My initial attempt would use the lipschitz (or Gronwalls Lemma?) However, I am unsure how to finish the problem.  

First we know that 
$$ |\frac{d}{dx} G(x)| = |2x| = 2|x|$$
but when $x \geq 1$ you know that  $|x| \leq |x|^2$, moreover this implies that  it is always true that $|x| \leq |x|^2 +1$  but we know that $ \frac{1}{c}|\frac{d}{dx} G(x)| =  |x|$ ( in our case the constant $1/c =1/2$), so from this we get:
$$ \frac{1}{c}  |\frac{d}{dx} G(x)| = |x| \leq x^2+1 = G(x)$$
$$ \rightarrow  |\frac{d}{dx} G(x)| \leq c G(x) $$
Hope this helps.
