Analysis problem: Show that$ f(x)$ is less or equal than$ g(x)$ Analysis problem:
Let $f$ and $g$ be differentiable on $ \mathbb R$. Suppose that $f(0)=g(0)$ and that $f' (x)$ is less or equal than $g' (x)$ for all $x$ greater or equal than $0$ Show that $f(x)$ is less or equal than$g(x)$ for all $x$ greater or equal than $0$.
Is my proof correct?
I am trying to use the Generalized Mean Value Theorem:

As $f$and g are differentiable on$ \mathbb R$, $f$ and $g$ are continuous on $ \mathbb R$ and we can use the Generalized Value Theorem. Using the starting condition $f(0)=g(0)$, we have that for any b that is greater than$ 0$, exist a $c$ element of $(0,b)$ such that
$f' (c) g(b) =  g' (c) f(b)$
By the starting conditions,
$f' (x) $is less or equal than$g' (x)$ for all $x$ greater or equal than $0$
Therefore, $f(b)$ is less or equal than $g(b)$ for any b element of $(0, b)$
As $b$ is any number bigger than$ 0$
$f(x)$ is less or equal than $g(x)$ for any $x$ greater or bigger than$ 0$. Q.E.D.

I don't really see how you conclude from 
$
f'(c)g(b)=g'(c)f(b)
$
and $f'(c)\leq g'(c)$ that $f(b)\leq g(b)$. For instance, $0<2$ and $0\times (-1) = 2\times 0$, but you cannot conclude that $0\leq-1$. 
As mentioned by dxiv, a proof can be achieved by using the Mean Value Theorem, applied to the function $g-f$: given any $x\geq0$, there exists $c$ with $0\leq c\leq x$ and $$g(x)-f(x)=g(x)-f(x)-(g(0)-f(0))=(g'(c)-f'(c))\,(x-0)\geq0.$$
It is also worth mentioning that the "direct way":
$$
f(x)=\int_0^xf'(t)\,dt\leq\int_0^xg'(t)\,dt=g(x)
$$
doesn't work in general, because it is not true in general that $f(x)=\int_0^xf'(t)\,dt$; this requires $f$ to be absolutely continuous (see also here and here).
