Show that the origin is an attractor point. $F'(0)<1$ Let $F:[0,1] \rightarrow [0,1]$ belong to $C^2([0,1])$ and verifies $F'(x)>0$ and $F''(x)>0$ for all $x \in (0,1)$.
Show that $F'(0) < 1$ (that the origin is an attractor point)
What I've come so far:
Expand $F'(x)$ through Taylor series around $0$:
$$F'(x) = F'(0)+F''(0)(x-0)$$
$$ F''(0) = \frac{F'(x)-F'(0)}{x}$$
Then if $F''(x)>0$ $\forall x \in (0,1)$:
$$\frac{F'(x)-F'(0)}{x} > 0$$
$$F'(x) > F'(0)\space \space\forall x \in (0,1)$$
Assume $F'(0) \geq 1$ then $F'(x) > 1 \space \space \forall x \in (0,1)$
Expand $F(x)$ through Taylor series around $0$:
$$ F(x) = F(0)+F'(0)(x-0)$$
$$F'(0) = \frac{F(x)-F(0)}{x} > 1$$
$$F(x) > x + F(0)$$
But since $F(x)$ must be bounded below by $x+F(0)$ and $F(x) \in [0,1]$, $x+F(0) \neq 0$ it would imply that $F(x) \in (0,1]$, so $F'(0)>1$
I'm not sure about this last argument which I think it's incorrect since assuming that $F'(0)<1$ I would get an upper bound of $F(x)$ with which I could use the same argument to refuse that $F'(0)<1$...

As mentioned in the comments, 
$$
F'(x) = F'(0)+F''(0)(x-0)
$$
is wrong (unless $F'$ is linear). What you have (from the mean-value theorem) is that
$$
F'(x) = F'(0)+F''(c)(x-0)
$$
for some $c \in (0,x)$. For the same reason,
$$
F(x) = F(0)+F'(0)(x-0)
$$
is not correct in general.

But you can proceed as follows: Taylor's formula for $F$ on the interval $[0, 1]$ gives
$$
 1 \ge F(1) = F(0) + (1-0) F'(0) + \frac 12 (1-0)^2 F''(c) 
\ge F'(0) + \frac 12  F''(c) > F'(0)
$$
for some $c \in (0,1)$.
Alternatively, use the mean-value theorem to show that
$$
 1 \ge \frac{F(1)-F(0)}{1-0} = F'(c)
$$ for some $c \in (0,1)$, and that $F'$ is strictly increasing.
