Miklos Schweitzer 2014 Problem 8: polynomial inequality Look at problem 8 :

Let $n\geq 1$ be a fixed integer. Calculate the distance: 
  $$\inf_{p,f}\max_{x\in[0,1]}|f(x)-p(x)|$$ where $p$ runs over
  polynomials with degree less than $n$ with real coefficients and $f$
  runs over functions $$ f(x)=\sum_{k=n}^{+\infty}c_k\, x^k$$ defined on
  the closed interval $[0,1]$, where $c_k\geq 0$ and
  $\sum_{k=n}^{+\infty}c_k = 1.$

This is what I have so far.
Clearly for $n=1$, we have $1/2$.
I am conjecturing for $n>1$, we have $(n-1)^{(n-1)} / n^n$ or something similar to that? (just put $x^{(n-1)}$ and $x^n$ then use AM-GM). it's just weird that the pattern does not fit, so it's probably wrong. Any ideas?

Your inequality does not hold since $x^{n-1}$ is not the best approximation polynomial of $x^n$ with respect to the uniform norm. By Chebyshev's theorem we have that if $p(x)$ is the best approximation polynomial for $f(x)$, then  $f(x)=p(x)$ holds for $\partial p+1$ points in $[0,1]$.
For instance, if $f(x)=x^n$ and $p(x)$ is the Lagrange interpolation polynomial with respect to the points $x=\frac{k}{n}$ for $k=1,2,\ldots,n$, since $f^{(n)}(x)=n!$ we have:
$$\|f(x)-p(x)\|_{\infty} = \left\|\prod_{k=1}^{n}\left(x-\frac{k}{n}\right)\right\|_{\infty}=\frac{n!}{n^n}$$
that is below your bound for any $n\geq 4$.
We can improve this bound by choosing our interpolation points more carefully: by selecting Chebyshev nodes, for instance: $x_k=\cos^2\frac{\pi(2k-1)}{4n}$ for $k=1,\ldots,n$.
In order to find the best approximation polynomial of $x^n$, have a look at the following answer of Noam Elkies on MO: https://mathoverflow.net/questions/70440/uniform-approximation-of-xn-by-a-degree-d-polynomial-estimating-the-error .
Since $\|T_n(2x-1)\|_\infty=1$, with the best choice for the interpolation nodes we have that the uniform error in approximating $x^n$ is always greater than $\color{red}{\frac{2}{4^n}}$.
Since for every function in our class we have $\frac{f^{(n)}(\xi)}{n!}\geq 1$ for any $\xi\in[0,1]$, $f(x)=x^n$ is the easiest function to approximate, and:
$$\inf_{p,f}\|f-p\|_{\infty}=\color{red}{\frac{2}{4^n}}.$$
