Your proof provides an interesting use of Hanson's lcm inequality result and appears to be basically correct, but there are few small mistakes & other issues.
In your step (5), you wrote
... $v_p({{3x}\choose{2x}}) = \sum\limits_{i \le \log_p(3n)} \left\lfloor\frac{3x}{p^i}\right\rfloor - \left\lfloor\frac{2x}{p^i}\right\rfloor - \left\lfloor\frac{x}{p^i}\right\rfloor$ which equals $1$ or $0$ for each $i$ ...
First, in the summation part of $i \le \log_p(3n)$, the $n$ I believe is supposed to be $x$. Also, in something like a math journal, the statement "equals $1$ or $0$ for each $i$" may be obvious & sufficient, but I don't believe it is for here since, I at least, didn't initially know if it was always true. FYI, here is what I did to confirm this. I let $d = p^i$ and $x = md + r$ for $m,d \in \mathbb{N}^0$ and $0 \le r \lt d$. Then,
\begin{align}
\left\lfloor \frac{3x}{d} \right\rfloor - \left\lfloor \frac{2x}{d} \right\rfloor - \left\lfloor \frac{x}{d} \right\rfloor & = 3m + \left\lfloor \frac{3r}{d} \right\rfloor - 2m - \left\lfloor \frac{2r}{d} \right\rfloor - m - \left\lfloor \frac{r}{d} \right\rfloor \\
& = \left\lfloor \frac{3r}{d} \right\rfloor - \left\lfloor \frac{2r}{d} \right\rfloor - \left\lfloor \frac{r}{d} \right\rfloor \\
& = \left\lfloor \frac{3r}{d} \right\rfloor - \left\lfloor \frac{2r}{d} \right\rfloor\tag{1}\label{eq1}
\end{align}
Since $r$ and $d$ are positive, both terms are non-negative and $\left\lfloor \frac{3r}{d} \right\rfloor \ge \left\lfloor \frac{2r}{d} \right\rfloor$, so \eqref{eq1} is non-negative. Since $\left\lfloor \frac{3r}{d} \right\rfloor \le 2$, the only way \eqref{eq1} can be anything other than $0$ or $1$ would be for $\left\lfloor \frac{3r}{d} \right\rfloor = 2$ (which requires $\frac{2d}{3} \le r \lt 1$) and $\left\lfloor \frac{2r}{d} \right\rfloor = 0$ (which requires $0 \le r \lt \frac{d}{2}$). Since there is no overlap between those $2$ regions, the result of \eqref{eq1} must always be $0$ or $1$.
Next, you state
If a prime $p > \sqrt{3x}$, then $v_p({{3x}\choose{2x}}) = 1$ and $p | \text{lcm}(\frac{3x}{2})$
For $x = 5$, note that $p = 5 \gt \sqrt{15}$, but ${{3x}\choose{2x}} = {{15}\choose{10}} = 3 \times 7 \times 11 \times 13$, so $v_p({{3x}\choose{2x}}) = 0$ in this case. A correct statement to make would be something like $v_p({{3x}\choose{2x}}) \leq 1$. Note this doesn't really affect your argument and it actually strengthens it slightly as you want to have a sufficiently small upper bound.
With the next statement of
If a prime $p \le \sqrt{3x}$, then $v_p({{3x}\choose{2x}}) \le \log_p(3x) \le v_p(\frac{3x}{2})+1$
first note I believe you mean the last part to be $v_p\left(\text{lcm}\left(\frac{3x}{2}\right)\right) + 1$ instead. However, for $p = 2$ and $x = 4$, then $\log_p(3x) = \log_2(12) \gt 3$, but $v_p\left(\text{lcm}\left(\frac{3x}{2}\right)\right) + 1 = v_2(60) + 1 = 2 + 1 = 3$. A correct statement could use the fact that $v_p$ is always an integer, so you can just use the integer component of the log to get $v_p({{3x}\choose{2x}}) \le \left\lfloor \log_p(3x) \right\rfloor \le v_p\left(\text{lcm}\left(\frac{3x}{2}\right)\right) + 1$.
Finally, for the second part of (7), you wrote
and for $x \ge 246$, $6 > (3^{3/2})(3^{\sqrt{3x+2} - \sqrt{3x}})\left(\dfrac{x+1}{x}\right)$
I don't see how you get this. Your step (6) states that $\dfrac{6^x}{x} < 3^{3x/2}3^{\sqrt{3x}}$. If you multiply by $x$ and take the $x$'th root of both sides, you get $6 \lt 3^{3/2}3^{\sqrt{3/x}}x^{1/x}$. I suggest not only showing how you got your result, but also how you proved it to be always be true for $x \ge 246$.
Here is how I would prove that (6) is not true for $x \ge 246$. First, note that since logarithms, including the natural logarithm, are a strictly increasing function for positive numbers, so $a \ge b \iff \log a \ge \log b$. Thus, take the natural logarithm of both side of (6) and move the right side to the left side to get the following function
\begin{align}
f(x) & = \ln(6)x - \ln(x) - \frac{3\ln(3)}{2}x - \ln(3)\sqrt{3}\sqrt{x} \\
& = \left(\ln(6) - \frac{3\ln(3)}{2}\right)x - \ln(3)\sqrt{3}\sqrt{x} - \ln(x) \tag{2}\label{eq2}
\end{align}
You've already shown that $f(246) \gt 0$. If can show that $f'(x) \ge 0$ for $x \ge 246$, then $f(x) \gt 0$ for all $x \ge 246$. Differentiating \eqref{eq2} gives
\begin{align}
f'(x) & = \left(\ln(6) - \frac{3\ln(3)}{2}\right) - \left(\frac{\ln(3)\sqrt{3}}{2}\right) \frac{1}{\sqrt{x}} - \frac{1}{x} \\
& = \frac{1}{x}\left(\left(\ln(6) - \frac{3\ln(3)}{2}\right)x - \left(\frac{\ln(3)\sqrt{3}}{2}\right)\sqrt{x} - 1\right) \tag{3}\label{eq3}
\end{align}
Note that
$$g(x) = \left(\ln(6) - \frac{3\ln(3)}{2}\right)x - \left(\frac{\ln(3)\sqrt{3}}{2}\right)\sqrt{x} - 1 \tag{4}\label{eq4}$$
is a quadratic equation in $\sqrt{x}$. As the coefficient of $x$, i.e., $\ln(6) - \frac{3\ln(3)}{2} = 0.143841\ldots \gt 0$, the value will always be positive for large enough $x$. Let $y = \sqrt{x}$ to transform \eqref{eq4} to
$$h(y) = \left(\ln(6) - \frac{3\ln(3)}{2}\right)y^2 - \left(\frac{\ln(3)\sqrt{3}}{2}\right)y - 1 \tag{5}\label{eq5}$$
Note that $h(\sqrt{246}) = 19.462358\ldots$. Also,
$$h'(y) = 2\left(\ln(6) - \frac{3\ln(3)}{2}\right)y - \left(\frac{\ln(3)\sqrt{3}}{2}\right) \tag{6}\label{eq6}$$
Since $h'(\sqrt{246}) = 3.560690\ldots$ and $h'(y)$ is a strictly increasing linear function, this shows that, in summary, $f(x)$ in \eqref{eq2} is always positive for $x \ge 246$, which confirms what you're trying to prove.
Best Answer
It's kind of a weird way of saying it. $$\mathrm{lcm}(P(x),Q(x)) = \frac{P(x)\cdot Q(x)}{\gcd(P(x),Q(x))}$$
Where $$\gcd(P(x),Q(x))$$ is just the product of the terms that appear in both factorizations.
For instance $$\mathrm{lcm}(((x+1)(x+1)(x-1)),((x+1)(x-1)(x-1)))=\frac{(x+1)^3(x-1)^3}{(x+1)(x-1)}=(x+1)^2(x-1)^2 $$