Consider a polynomial $P\_n(x)\in\mathbb{R}[x]$, of degree $n\geq1$, of the form
$$P\_n(x)=c\_0+c\_1x+c\_2x^2+\cdots+c\_{n-1}x^{n-1}+x^n.$$
To illustrate the question, take $P\_1(x)=c\_0+x$ so that $P\_1(0)=c\_0$ and $P\_1(1)=c\_0+1$. If $\vert c\_0\vert<\frac12$ then $\vert c\_0+1\vert\geq1-\vert c\_0\vert>1-\frac12=\frac12$. That means, $\pmb{\max}\{\vert P\_1(0)\vert,\vert P\_1(1)\vert\}\geq\frac12$.


In general,



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> **QUESTION.** is this true?
>  $$\pmb{\max}\{\vert P\_n(0)\vert,\vert P\_n(1)\vert,\vert P\_n(2)\vert,\dots,\vert P\_n(n)\vert\}\geq\frac{n!}{2^n}.$$
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Largest absolute value of a polynomial of degree $n$ on $\{0,1,\ldots,n\}$
You can write your polynomial as
$$ P(x) = \sum\_{k = 0}^n P(k) L\_{n,k}(x) ,$$
where $L\_{n,k}$ are the Lagrange interpolation polynomials with nodes at $0, 1, \ldots, n$. Note that $(-1)^{n - k} L\_{n,k}$ has positive coefficient at $x^n$. Thus, with the constraint $$\max\{|P(k)| : k = 0, 1, \ldots, n\} \leqslant 1,$$ the largest possible value of the coefficient of $P$ at $x^n$ is attained by
$$ \bar P(x) = \sum\_{k = 0}^n (-1)^{n - k} L\_{n,k}(x) , $$
and the coefficient of $\bar P$ at $x^n$ is equal to
$$ \sum\_{k = 0}^n \prod\_{\substack{0 \leqslant j \leqslant n \\ j \ne k}} \frac{1}{|k - j|} \, .$$
In other words, if the coefficient at $x^n$ is to be equal to $1$, the least possible value of $\max\{|P(k)| : k = 0, 1, \ldots, n\}$ is
$$ \biggl(\sum\_{k = 0}^n \prod\_{\substack{0 \leqslant j \leqslant n \\ j \ne k}} \frac{1}{|k - j|} \biggr)^{-1} .$$
It remains to note that
$$ \sum\_{k = 0}^n \prod\_{\substack{0 \leqslant j \leqslant n \\ j \ne k}} \frac{1}{|k - j|} = \sum\_{k = 0}^n \frac{1}{k! (n - k)!} = \frac{1}{n!} \sum\_{k = 0}^n \binom{n}{k} = \frac{2^n}{n!} \, .$$


