Suppose that $F'(x)\leq G'(x)$ for all $x \in \mathbb{R}$. Then $F(x)\leq G(x)$ for all $x \in \mathbb{R}$. Suppose that $F'(x) \leq G'(x)$ for all $x \in \mathbb{R}$. Then $F(x) \leq G(x)$ for all $x \in \mathbb{R}$. Prove or disprove. I came up with counterexample that  $\cos(x) \leq \cos(x)+1 $ for all $x$, but this does not imply that their antiderivatives $\sin(x) \leq \sin(x)+x $, if we choose any negative $x$ this will not hold. Am I right or wrong?
 A: Simpler counter-example: $F(x) = 1$, $0 = G(x)$ :)
However it's true that: assuming $F(x_0) \le G(x_0)$ and $F'(x) \le G'(x)$ for $x \ge x_0$ we have $F(x) \le G(x)$ for all $x \ge x_0$.
A: Take any bounded function as $g(x)$ and take $f(x)$ as the function $g(x)$ shifted upward high enough that $g(x) < f(x)$ (we can do this if $g$ is bounded).  Then since shifting upward doesn't change the derivative (i.e., if $f(x) = g(x) + c$ then $f'(x) = g'(x)$), we have $f'(x) \leq g'(x)$ but $f(x) > g(x)$ for all $x$.
Here is an easy example of two lines which have the same slope (note that these aren't bounded as in my method above):
Let $f(x) = 3x + 1$ and $g(x) = 3x$.  (You should be thinking of these geometrically as the lines they are.)
Then $f'(x) = g'(x) = 3$, so $f'(x) \leq g'(x)$.
But it should be clear that $f(x) > g(x)$ for all $x$, so we don't have $f(x) \leq g(x)$.
A: Best you can say is $G'(x) \leq F'(x) \rightarrow (F(x)-G(x))' \geq 0 $ so that $F(x) \geq G(x) $.
Now this is true if we know if $F(x) > G(x)$ for just one $x>0$.
If $F(x)>G(x)$ for some x, then $F(x)-G(x)>0$ , then use that $(F(x)-G(x))'>0$, so that $F(x)-G(x) \geq 0$ is nondecreasing for $x>0$, so that $F(x)-G(x) \geq 0$.
A: The claim is wrong. Consider example like: 
$F(x)=x, G(x)=2x-1$
So $F'(x)=1, G'(x)=2$, then 
For all $x\in \mathbb{R}, F'(x)<G'(x)$, but 
$\forall x<1,F(x)>G(x)$ and  $\forall x>1,F(x)<G(x)$
