Find the largest leading coefficient of $P(x)$ Let $ P(x)=Ax^4+Bx^3+Cx^2+Dx+E$. Find the largest value of $A$ which satisfies $0 \leq P(x) \leq 1$ for all $x \in [-1,1]$.
I have no idea on how to start. It looks quite weird(I have never seen such type of problems). How do I start? Thanks.

First notice that we can assume that $P(x)$ has a root in $[-1,1]$ and that exists $c\in [-1,1,]$ such that $P(c)=1.$
Suppose now $0\le P(x)\le1$ for every $x$ in $[-1,1]$, then 
$$0\le \frac{P(x)+P(-x)}{2}\le1.$$
We have that $\frac{P(x)+P(-x)}{2}=Ax^4+Cx^2+E$, so we can suppose $B=D=0$ and $0\le E\le1$.
If $C\ge0$ then we must have $A\le1$. So suppose now $C<0$. Let's now consider the case when $P(\pm1)\neq 0$, so the root of $P(x)$ must be inside $(-1,1)$ and so it must be a local minimum. 
$$P'(x)=2x(2Ax^2+C)$$
so the minima are at $x^2=-C/2A$ and setting $P(-C/2A)=0$ we get
$$A=\frac{C^2}{4E}.$$
Now we use the fact that exists $c\in [-1,1,]$ such that $P(c)=1$ and suppose $c\neq \pm1$. In this case $c$ must be a local maximum, so $c=0$ and so $E=1$. Hence $A=C^2/4$ and finally using the fact that $P(1)=A+C+E=\le1$ we get 
$$A\le4.$$
When $A=4$ we obtain the polynomial
$$P(x)=4x^4-4x^2+1.$$
We should check the case when $c=\pm1$, that is when $A+C+E=1$ and $A=C^2/4E$, but I found (after some computation) that the maximal $A$ is again $A=4$, when $E=1$.
