Reasoning about least common multiples and their relationship to binomial coefficients I have been fascinated by this inequality which I believe is valid:
Let: 


*

*$x \ge n$ be integers

*lcm$(a,b,c)$ be the least common multiple of integers $a, b, c$
$$\text{lcm}(x+1, x+2, \dots, x+n) \le {{x+n}\choose{n}}\text{lcm}(1,2,\dots,n)$$
Here's my reasoning:

  
*
  
*lcm$(x+1,x+2,\dots,x+n) < \frac{(x+n)!}{x!}$
  
*If a prime $p > n$ and $p | \dfrac{(x+n)!}{x!}$, then $p | {{x+n}\choose{n}}$ since $p \nmid n!$
  
*If a prime $p \le n$ with $p^t > n$ and $p^t | \dfrac{{x+n}!}{x!}$, then there exists $c$ where $p^{t-c} | {{x+n}\choose{n}}$ where $p^c | \text{lcm}(1, 2, \dots, n)$ and $p^{c+1} \nmid \text{lcm}(1,2,\dots,n)$
  
*It follows that lcm$(x+1, x+2, \dots, x+n) \le {{x+n}\choose{n}}\text{lcm}(1,2,\dots,n)$

I am wondering if the following extension holds:
$$\text{lcm}(\text{lcm}(x-n+1, x-n+2,\dots,x),\text{lcm}(x+1, x+2, \dots, x+n)) = \text{lcm}(x-n+1, x-n+2, \dots, x+n) \le {{x}\choose{n}}{{x+n}\choose{n}}\frac{(n!)^2}{(2n)!}\text{lcm}(1,2,\dots,2n)$$
Here's my thinking for this extension:
(1) lcm$(x-n+1, x-n+2, \dots, x+n) \le {{x+n}\choose{2n}}\text{lcm}(1,2,\dots,2n)$
(2) ${{x+n}\choose{2n}} = \dfrac{(x+n)!}{(x-n)!(2n)!} = \left(\dfrac{x!}{(x-n)!(n!)}\right)\left(\dfrac{(x+n)!}{(x!)(n!)}\right)\left(\dfrac{(n!)(n!)}{(2n)!}\right) = {{x}\choose{n}}{{x+n}\choose{n}}\frac{(n!)^2}{(2n)!}$
Does this extension hold?  Is there a standard way to represent this relationship?  I am especially interested if there is a related statement in terms of Stirling Approximation and the second Chebyshev function.
 A: Your inequality, i.e.,
$$\operatorname{lcm}(x + 1, x + 2, \, \ldots, \, x + n) \le {{x + n}\choose{n}}\operatorname{lcm}(1, 2, \, \ldots, \, n) \tag{1}\label{eq1A}$$
is true, actually for all integers $n \ge 1$ and $x \ge 0$, not just for where $x \ge n$.
However, there are several problems with your attempted proof. A fairly minor one is with your first bullet point. Note both sides of the inequality are actually always equal for $n = 1$ and $n = 2$, and equal for $n = 3$ when $x + 1$ is odd, so you should be using $\le$ instead of $\lt$.
Regarding your third bullet point, as pointed out in mathlove's question comment, there's a counter-example. In your comment reply, you suggested defining $t$ as being $p^t \mid \operatorname{lcm}(x + 1, \, \ldots \, x + n)$ instead, but you don't explain how to modify the rest of the bullet point accordingly.
With your extension inequality, if you were trying to use the result of your first inequality, due to its restriction of the first value in the $\operatorname{lcm}$ being $\ge n + 1$, then your extension would be limited to where $x - n + 1 \ge n + 1 \implies x \ge 2n$. Otherwise, your thinking is correct. Actually, your extension is true for all $x - n + 1 \ge 1 \implies x \ge n$.

Here is one way to prove \eqref{eq1A} is true for all integers $n \ge 1$ and $x \ge 0$. First, for somewhat simpler algebra, let
$$a = \operatorname{lcm}(x + 1, \ldots, \, x + n), \; b = {{x + n}\choose{n}}, \; c = \operatorname{lcm}(1, \, \ldots, \, n) \tag{2}\label{eq2A}$$
Note $b$ can also be expressed as
$$b = \frac{(x + 1)(x + 2)\, \ldots \, (x + n)}{n!} = \frac{\prod_{i=1}^{n}(x + i)}{n!} \tag{3}\label{eq3A}$$
Using the prime factorization exponents based on the definition of $\operatorname{lcm}$, and the $p$-adic order function, then for any prime $p$,
$$d = \nu_{p}(a) = \max_{1 \, \le \, i \, \le n}(\nu_{p}(x + i)) \tag{4}\label{eq4A}$$
$$e = \nu_{p}(b) \tag{5}\label{eq5A}$$
$$f = \nu_{p}(c) = \max_{1 \, \le \, i \, \le n}(\nu_{p}(i)) \tag{6}\label{eq6A}$$
By proving
$$d \le e + f \iff d - f \le e \tag{7}\label{eq7A}$$
for all primes $p$, then \eqref{eq1A} will be true.
Note \eqref{eq6A} gives
$$p^f \le n \lt p^{f+1} \tag{8}\label{eq8A}$$
Since $a$ is the product of at least $p^f$ consecutive integers, they represent a complete residue system of integers modulo $p^f$, so one or more of them must be $\equiv 0 \pmod{p^f}$, which means
$$d \ge f \tag{9}\label{eq9A}$$
If $d = f$, then \eqref{eq7A} gives $e \ge 0$, which is always true. Consider instead that $d \gt f$. Due to the upper bound in \eqref{eq8A}, there's only one $1 \le j \le n$ where $\nu_{p}(x + j) = d$. Also, let $1 \le k \le n$ be any of the one or more values where $\nu_{p}(k) = f$. Next, from \eqref{eq5A},
$$\begin{equation}\begin{aligned}
e & = \left(\sum_{i=1, \, i \neq j}^{n}\nu_{p}(x + i) + d\right) - \left(\sum_{i=1, \, i \neq k}^{n}\nu_{p}(i) + f\right) \\
e & = (d - f) + \sum_{i=1, \, i \neq j}^{n}\nu_{p}(x + i) - \sum_{i=1, \, i \neq k}^{n}\nu_{p}(i) \\
e - (d - f) & = \sum_{i=1, \, i \neq j}^{n}\nu_{p}(x + i) - \sum_{i=1, \, i \neq k}^{n}\nu_{p}(i)
\end{aligned}\end{equation}\tag{10}\label{eq10A}$$
Thus, we want to prove the right side of the last line above is non-negative. For some positive integer $m$, we have
$$x + j = mp^{d} \implies \color{blue}{x - mp^{d}} = j \tag{11}\label{eq11A}$$
Define a new positive starting value of $x$ as
$$x_{1} = p^{f+1} - (\color{blue}{x - mp^{d}}) \implies x_1 + j = p^{f 
 + 1} \tag{12}\label{eq12A}$$
Note
$$\begin{equation}\begin{aligned}
x + i & = x + j + (i - j) = mp^d + (i - j) \\
x_1 + i & = x_1 + j + (i - j) = p^{f+1} + (i - j)
\end{aligned}\end{equation}\tag{13}\label{eq13A}$$
Since $\nu_p(x + i) \le f$ for all $1 \le i \le n, \; i \neq j$, then we also have for all such $i$ that
$$\nu_{p}(i - j) = \nu_{p}(x + i) = \nu_{p}(x_1 + i) \tag{14}\label{eq14A}$$
Thus, the value of the right side of the last line of \eqref{eq10A} is the same if $x$ is replaced by $x_1$. Next,
$$\begin{equation}\begin{aligned}
e_1 & = \nu_{p}\binom{x_1 + n}{n} \\
& = \left(\sum_{i=1, \, i \neq j}^{n}\nu_{p}(x_1 + i) + (f + 1)\right) - \left(\sum_{i=1, \, i \neq k}^{n}\nu_{p}(i) + f\right) \\
& = 1 + \sum_{i=1, \, i \neq j}^{n}\nu_{p}(x_1 + i) - \sum_{i=1, \, i \neq k}^{n}\nu_{p}(i)
\end{aligned}\end{equation}\tag{15}\label{eq15A}$$
Thus, we just need to prove that $e_1 \ge 1$. This can be shown using Kummer's theorem which states $e_1$ is the number of carries when $n$ is added to $x_1$ in base $p$. Since $2p^{f + 1} \gt x_1 + n \ge x_1 + j = p^{f+1}$, then $x_1 + n$ has a $1$ for the $p^{f+1}$ digit in base $p$. However, $p^{f+1} \gt x_1$ and $p^{f+1} \gt n$, so both have $0$ in that digit, which means there must be at least one carry in the addition, giving that $e_1 \ge 1$, so the left side of \eqref{eq10A} is $\ge 0$. Thus, this shows \eqref{eq7A} is true in this case. Since we've now covered all of the cases, then \eqref{eq1A} is also true.
