If $p\in\Bbb Z[X]$ show that: $ \max\limits_{x\in [0,1]}\left|p(x) \right| > \frac{1}{e^{n}}. $ 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.
 A: 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 $$ \color{blue}{M = \max_{x\in [0,1]}\left|p(x)\right| \ge \frac{1}{e^n }}$$
A: I have to remark that the given inequality is fairly simple to prove if we replace $e$ with $4$.
Since the shifted Legendre polynomials provide an orthogonal base of $L^2(0,1)$ with respect to the standard inner product, we have
$$ \min_{\deg q<n}\int_{0}^{1}\left(x^n+q(x)\right)^2\,dx = \frac{1}{(2n+1)\binom{2n}{n}^2} $$
hence $p(x)\in\mathbb{Z}[x]$ and $\deg p=n$ imply
$$ \max_{x\in[0,1]}\left| p(x) \right| \geq \frac{1}{\binom{2n}{n}\sqrt{2n+1}} >\frac{1}{4^n}.$$
A: Not a full answer, but adding to Colm's answer, we begin by noting that if the condition is being violated for any polynomial P(x), P(0) must be zero as must be P(1).
This is because both P(0) and P(1) are necessarily integral and will always exceed e^(-n) where n is as defined.
Hence, such a polynomial will always be expressible as x(x-1)g(x). By induction, we see that if |g(x)| reaches its maxima in a range lying in [1/2-k,1/2+k] such that the value of k allows x(x-1) to exceed e^(-2) in absolute value, we will have our contradiction and the solution.
A: Not a full answer Just a (fairly obvious) idea for tackling this.
The case for $n = 1$ is just short of trivial. Let the polynomial be $ax + b$. Then if $|b| > \frac{1}{e}$, we can choose $x = 0$ and we're done. Else $|b| \leq \frac{1}{e}$. So choosing $x = 1$ we get $|ax + b| = |a + b| > 1 - \frac{1}{e} > \frac{1}{e}$. The last inequality chain holds because $a$ must be an  integer.
Idea from here: Induction on $n$. The inductive hypothesis gives us two results:


*

*The "sub-polynomial" of degree $n - 1$ reaches a value greater than $\frac{1}{e^{n - 1}}$ at some stage

*The derivative, which is of degree $n - 1$ reaches a value greater than $\frac{1}{e^{n - 1}}$ at some stage.


Maybe also the Taylor series expansion for $e^{-nx}$ would be useful here, because then we're comparing polynomials with polynomials.
