TITLE: If $p\in\Bbb Z[X]$ show that: $ \max\limits_{x\in [0,1]}\left|p(x) \right| > \frac{1}{e^{n}}. $
QUESTION [31 upvotes]: This is problem 10 from the International Mathematical Competition for University Students of 2015, from day 2, in Bulgaria. I think it is an interesting problem!

Let $n$ be a positive integer, and $p(x)$ be a polynomial of degree $n$ with integer coefficients. Prove that
  $$
\max_{x\in [0,1]}\left|p(x) \right| > \frac{1}{e^{n}}.
$$

Proposed by Géza Kós, Eötvös University, Budapest.

REPLY [7 votes]: Let $ M = \max_{x\in [0,1]}\left|p(x) \right| $ we know that 
$$ M = \max_{x\in [0,1]}\left|p(x)\right| =\lim_{j\to \infty}\left(\int_0^1 \left|p(x)\right|^j dx\right)^{\frac{1}{j}}=\lim_{j\to \infty}\left(\int_0^1 \left|p(x)\right|^{2j} dx\right)^{\frac{1}{2j}}$$
We then consider
$$ M_{j} =\int_0^1 \left|p(x)\right|^{2j} dx$$

since $\deg p= n$ then it follows that , $\deg p^{2j}= 2jn$. so, If we Can write it as, 
  $$p^{2j}(x) = \sum_{k=0}^{2jn} a_{j,k}x^k$$

Therefore, 
$$ M_{j} =\int_0^1 \left|p(x)\right|^{2j} dx \ge\left|\int_0^1 p(x)^{2j} dx \right| = \left|\sum_{k=0}^{2jn} \frac{a_{j,k}}{i+1}\right|$$
Let $\color{blue}{\operatorname{lcm}(1,2,3,\cdots,  2jn+1)}$ be the least common factor of $(1,2,3,\cdots,  2jn)$. Then since $a_{k,i}\in \Bbb Z$ we have, 
$$ M_{j}\ge \left|\sum_{k=0}^{2jn} \frac{a_{j,k}}{i+1}\right|\ge \frac{1}{lcm(1,2,3,\cdots,  2jn+1)}$$
We get from this that, 
$$\color{red}{\log (\color{blue}{\operatorname{lcm}(1,2,3,\cdots,  2jn+1)})\sim 2jn+1}$$
hence for $\varepsilon>0$ and for $j$  large enough we have, 
$$\color{blue}{\operatorname{lcm}(1,2,3,\cdots,  2jn+1)}\le \exp((1+\varepsilon)(2jn+1) ) $$
which implies that 
$$ \left(\int_0^1 \left|p(x)\right|^{2j} dx\right)^{\frac{1}{2j}} = M_j^{1/2j} \ge \left(\frac{1}{\color{blue}{\operatorname{lcm}(1,2,3,\cdots,  2jn+1)}}\right)^{\frac{1}{2j}} \ge \frac{1}{\exp((1+\varepsilon)(n+\frac{1}{2j}) ) }$$
Taking $j\to \infty$ we get 
$$ M = \max_{x\in [0,1]}\left|p(x)\right| = \lim_{j\to \infty}\left(\int_0^1 \left|p(x)\right|^{2j} dx\right)^{\frac{1}{2j}} \ge \frac{1}{\exp((1+\varepsilon)n ) }~~~\forall ~~\varepsilon>0$$
That is $$ \color{blue}{M = \max_{x\in [0,1]}\left|p(x)\right| \ge \frac{1}{e^n }}$$