how to prove that the sum of the power two of three reals is less than 5? $a, b$ and $c$ are three real number such that: $(\forall x \in [-1,1]): |ax^2+bx+c|\leq 1$ .

Prove that:

1)   $|c|\leq 1$ .

2) $-1\leq a+c\leq 1$ .

3)Deduce that : $a^2+b^2+c^2\leq 5$.

The first and second questions are easy to prove ( just take $x=0, -1$ or $1$ ... ), but the third one ?!!! I'm sure that the responce must be easy since it's a deduction. 

I need your help to solve it, thank you.
$x=0$ gives us $|c| \le 1$.  Similarly set $x = \pm1$ to get $|a \pm b+c| \le 1$.  So we have the inequalities:

$$-1 \le c \le 1 \iff -1 \le -c \le 1\tag{1}$$
$$-1 \le a+b+c \le 1 \iff -1\le -a-b-c \le 1 \tag{2}$$
$$-1 \le a-b+c \le 1 \iff -1 \le -a+b-c \le 1\tag{3}$$

Note the right side of the above three inequalities can be considered the "negatives" of the left sides.  These should be enough for us to prove all sorts of things.

Here, 
$$(2)-(1) \implies -2 \le a+b \le 2\implies (a+b)^2 \le 4$$
$$(3)-(1) \implies -2 \le a-b \le 2\implies (a-b)^2 \le 4$$

Now those two give $a^2+b^2=\frac12((a+b)^2+(a-b)^2) \le 4$.
We just need to add to this $|c| \le 1 \implies c^2 \le 1$ to get what we want.
I'm not sure how you're supposed to go straight from 1) and 2) to 3) without some additional steps.

The help I would offer is that it might help to strengthen 2) to the statement $|a+c| + |b| \le 1$ and work from there.