Showing that constant term of the polynomial satisfy the below inequality If a polnomial $R(x)$ of ninth degree satisfies $|a_9x^9 + a_8x^8+..a_0|$ $\leq$ $1$ $\forall$ x $\in$ $[-1,1]$ - {${0}$} , then show that $a_0$ satisfies $|a_0| \leq 1$. And does equality every achieved ?
This was originally from a problem:  Amongst all polynomials $p(x)=c_{0}+c_{1} x+\cdots+c_{10} x^{10}$ with real, coefficients satisfying $|p(x)| \leq|x|$ for all $x \in[-1,1]$. What is the maximum possible value of $\left(2 c_{0}+c_{1}\right)^{10 ?}$.          where its easy to see $c_0$ is $0$ when we put $x= 0$ , so we can write $p(x)$ = $x(R(x))$ and then $|p(x)| = |xR(x)|$ $\leq |x|$ , cancelling mod x for x $\neq$ 0 we get the new question which i posted above.

My method was :
We can let $x$ to approach zero and put it in the expression we get for very very close values to $0$ that $a_0 <1$ , it seems like this is sufficient to conclude the result but not sure if this is correct way to judge it or not.    ( That is $|R(0^+)|$ =  $|a_9(0^+)^9 .... +a_0 |$  , all the high order terms leaving the $a_0$ can be assumed to approach zero , hence $|a_0| < 1 $)

 A: Assume that $a_0 > 1.$ The demonstration, if $a_0 < -1$ will be similar.
Let $f(x) = a_9x^9 + \cdots + a_1x^1.$
Let $M = \max(|a_9|, |a_8|, \cdots, |a_1|).$
Construct $g(x) = -(1) \times M \times (x^9 + x^8 + \cdots + x^1).$
Since you are permitted to choose $x > 0$ as small as you want, you can certainly find $x_0 > 0$ such that $g(x_0) > 1 - a_0.$  By the construction of the function $g$, this implies that $f(x_0) > 1 - a_0$.
This implies that $f(x_0) + a_0 > 1,$ which violates the constraint that $|f(x) + a_0| \leq 1,$ for all $x$ in the $[-1,1] - \{0\}$ interval.

Now, consider what happens if $a_0 = 1$.  Since it is assumed that $f(x)$ is a 9th degree polynomial, you must have that $a_9 \neq 0.$  Reading the coefficients $a_9, a_8, \cdots, a_1$, from left to right, let $a_k$ denote the rightmost non-zero coefficient.  For example, if $k = 5$, this implies that $a_4, a_3, a_2, a_1$ are all $0$.
First, I will present a hand-waving argument, that you can not have $a_0 = 1.$  Then, I will justify it.  Assume that $a_k > 0.$  As $x$ approaches $0$ from above (i.e. from the positive side), $a_kx^k$ will dwarf each of $a_9x^9, a_8x^8, \cdots, a_{k+1}x^{k+1}$.  This is because, you have total freedom to choose $x > 0$ as small as you want, and you have that $x^k$ is $\frac{1}{x}$ times greater than $x^{k+1}.$
This guarantees, that you will be able to find an $x_0 > 0$ such that $f(x_0) > 0.$  This will yield a contradiction, because, since $a_0 = 1,$ the result will be that $|f(x_0) + a_0| > 1.$

Formal argument:
With $a_0 = 1$ (rather than $-1$), assume that $a_k > 0.$  Here, the corresponding argument if $a_k < 0$ will be parallel, since you will be allowed to choose $x_0 < 0$ as large as you want.
Without loss of generality, $k < 9.$  That is, if $a_9 > 0$ is the rightmost non-zero coefficient, then you can obviously find $x_0$ such that $f(x_0) > 0.$
Let $M$ denote $~\displaystyle \max\left(\frac{|a_9|}{a_k}, \frac{|a_8|}{a_k}, \cdots, \frac{|a_{k+1}|}{a_k}\right).$
Choose $x_0 > 0$ small enough so that
$\displaystyle (9-k) \times \frac{1}{x_0} > M.$
This will guarantee that
$\displaystyle |a_r(x_0)^r| < \frac{a_k(x_0)^k}{9-k} ~: r \in \{9,8,\cdots,k+1\}.$
This implies that $f(x_0)$ is strictly greater than $0$.
Then, since $a_0 = 1,$ you will have the contradiction that $|f(x_0) + 1| > 1.$
