Find the maximum value of a constant $\psi$ for which the inequality $\psi |x - y| \leq |f(x) - f(y)|$ holds 
Let $n$ be a natural number not equal to $0$. Find the maximum value of the constant $\psi(n)$ for which for every polynomial $f(x)$, with $\deg f = n$, with the property that for every integer $x$, $f(x)$ is also an integer, satisfies the following relation:
$$\psi |x - y| \leq |f(x) - f(y)|$$
for every distinct integers for which $f(x) \neq f(y)$.

Attempt
I will prove that $\psi = 1$. For the polynomial $f(x) = x^n$ (obviously satisfies the property mentioned), and for $x = 1$, $y = 0$, we will get:
$$\psi|1 - 0| \leq |1 - 0|$$
so
$$\psi \leq 1$$
Now I shall prove that for every polynomial of every degree, $\psi = 1$ is a solution. I'm thinking somehow informal at the 'discrete' derivative, defined on discrete valued polynomials:
$$\frac{\delta P}{\delta x} = \lim_{\ell \to 1} \frac{P(x + \ell) - P(x)}{\ell}$$
And what we need to prove it's true that, for each polynomial of non-zero degree, $\left|\frac{\delta P}{\delta x}\right| > 1$ for every $x$, and this looks true, since induction principles say that 'polynomial function beats the linear/identity' (while considering only discrete/integer values).
Question:

How can I formalize my proof? Is it correct? Have you any other suggestions of how to approach this type of problem?

 A: We will show that $\psi(n) = \frac{1}{\text{lcm}(1,2,...,n)}$.
$\textbf{Background}:$
For a prime $p$ and a natural number $n$ we set
$M(p,n) = \max{\{j: p^j|n}\}$. I.e $M(3,15) = 1$.
$L(p,n) = \max{\{j: p^j \leq n\}}$. I.e $M(3,15) = 2$
$\text{lcm}(1,2,...n)$ is the least common multiple of the numbers $1,2,...,n$. I.e $\text{lcm}(1,2,3) = 6, \text{lcm}(1) = 1$.
$\textbf{Proof}:$
By translating the polynomial and making the constant term $0$, it suffices to show that whenever $f(y) \neq 0$ we have $$|f(y)| \geq \frac{1}{\text{lcm}(1,2,...,n)}|y|$$ whenever $f$ is a degree $n$ polynomial.
Set $$g_{n}(x) = \binom{x}{n} = \frac{x(x-1)...(x-n+1)}{n!}$$
Note by https://en.wikipedia.org/wiki/Integer-valued_polynomial that for a degree $n$ integer polynomial which vanishes at $0$, there exists constants $\lambda_{j} \in \mathbb{Z}$ so that
$$(1) \text{ }\text{ }\text{ }f(x) = \sum_{j=1}^{n}\lambda_{j}g_{j}(x).$$
We need to show for all $x$ that
$$\gcd_{1 \leq j \leq n}{g_{j}(x)} \geq \frac{|x|}{\text{lcm}(1,2,...,n)}.$$
As this will show whenever $f(x)$ is non-zero we have
$$|f(x)| \geq \frac{|x|}{\text{lcm}(1,2,...,n)}$$
when we consider $(1)$ and linear number theory.
Suppose $x \neq 0$ is of the form $bc$ where $\gcd(c,n!) = 1$, if some prime $p | b \Rightarrow p \leq n$ $\Rightarrow$ then $\gcd_{1 \leq j \leq n}{g_{j}(x)}$ is a multiple of $c$ and is a divisor of $x$ as $g_{1}(x) = x$. We have two cases
Case 1) Some prime $p |b$ satisfies $M(p,b)>L(p,n)$.
Set $M(p,b) = d$. so that $x = p^d\alpha$; note that
$$g_{j}(p^{d}\alpha) = \frac{(p^d\alpha)(p^d\alpha-1)...(p^d\alpha-j+1)}{j!}$$
So that
$$M(p, g_{j}(p^{d}\alpha)) = d + M(p,(j-1)!) - M(p,j!) = d - M(p,j) \geq d - L(p,n)$$
with equality iff $j \equiv 0 \mod p^{L(p,n)}$, $1 \leq j \leq n$. Thus $$M(p, \gcd_{1 \leq j \leq n}{g_{j}(x)}) = d - L(p,n) = M(p,b) - L(p,n).$$
Case 2) Some prime $p|b$ satisfies $M(p,b) \leq L(p,n)$
Note that $g_{p^{M(p,n)}}(x) \not \equiv 0 \mod p$ as
Set $M(p,n) = d$ so that $x = p^d \alpha$. Note
$$g_{p^{d}}(x) = \frac{(p^d\alpha)(p^d\alpha-1)...(p^d\alpha-p^d+1)}{(p^d)(p^d-1)...1}$$
Thus $$M(p,g_{p^{d}}(x) ) = M(p,p^d\alpha)+M((p^d-1)!) - M(p,p^d)- M(p,(p^d-1)!) = 0.$$
Thus $$p \not \div \gcd_{1 \leq j \leq n}{g_{j}(x)} $$.
Hence by the above cases, we have the equality
$$\gcd_{1 \leq j \leq n}{g_{j}(x)} = \prod_{p, p \text{ prime},\text{ } p|x,\text{ } 1\leq p \leq n,\text{ } M(p,x) \geq L(p,n)}p^{M(p,x)-L(p,n)}\prod_{p, p\text{ prime }p >n}p^{M(p,x)}$$
$$\geq \frac{|x|}{\prod_{p, p \text{ prime},\text{ } 1\leq p \leq n}p^{L(p,n)}} = \frac{|x|}{\text{lcm}(1,2,...,n)}$$
with equality iff $x$ is a multiple of $\text{lcm}(1,2,...,n)$. This shows that $\psi(n) \geq \frac{1}{\text{lcm}(1,2,...,n)}$. Now consider $y = \text{lcm}(1,2,...,n)$. Note $\text{gcd}_{1 \leq j \leq n}(g_{j}(y)) = 1$. Hence by $(1)$ there exists constants $\lambda_{j} \in \mathbb{Z}$ so that
$$\sum_{j=1}^{n}\lambda_{j}g_{j}(y) = 1$$ so that
$$|\sum_{j=1}^{n}\lambda_{j}g_{j}(y)-\sum_{j=1}^{n}\lambda_{j}g_{j}(0)| = 1 = \frac{1}{\text{lcm}(1,2,...,n)}|y|.$$
This finishes the proof.
A: Conceptually you are asking about a lower bound on the absolute value of the slope of non-horizontal secant lines between points on a curve that all have integer values of x.
I take it that you meant to include a requirement that all coefficients are also integers, since without such a requirement the only lower bound possible on the absolute value of the slopes of secant lines would be 0.
Under those requirements, taking the difference quotient [f(x+h)-f(x)]/h for an arbitrary polynomial function with integer coefficients, the coefficients of each term get multiplied by the order of the term, and the order of the term subsequently goes down by one. Since these new coefficients are each the product of two integers, these new coefficients are themselves integers. The result is that using only integer values of x.
Since this difference quotient represents the slope of the secant line, it is a polynomial function, and all its coefficients are integers, you can say that for any two integer values of x, the slope of the secant line must be an integer. If that slope is 0, then it fails to meet the stipulation that f(x)$\neq$f(y). Otherwise the absolute value of the slope is at least 1, since for any non-zero integer its absolute value is greater than or equal to 1.
A: I wasn't able to solve the problem but I have some ideas for a good start
Firstly, $\psi(1)$ is obviously $1$ and the for example we can take the polynomial $X+1$.
For $n=2$ things start to get weird. $\psi(2)=\frac{1}{2}$ because we can take the polynomial $\frac{X^2-X}{2}$ which satisfies the condition and for $X=1$ and $Y=-1$ we get that $\psi(2)=\frac{1}{2}$. This is actually easy to prove that it is minimal.
However for $n\geq 3$ things get too weird too work with so I think that we need to aproach this problem from another angle.
I hope that my ideas can help someone make a complete proof.
