Using Least common multiple to establish a lower bound for a ratio of primorials

Let $$x\#$$ be the primorial for $$x$$.

Let $$\text{lcm}(x)$$ be the least common multiple of $$\{1, 2, 3, \dots, x\}$$.

It occurs to me that for $$x \ge 4$$, it is straight forward to find a lower bound of $$\dfrac{(x^2+x)\#}{\left(\frac{x^2+x}{2}\right)\#}$$.

I find it interesting that an upper bound can lead to a lower bound.

Is my reasoning correct?

Here's my argument:

(1) $$\text{lcm}\left(\dfrac{x^2+x}{2}\right)\text{lcm}(x)\left(\dfrac{(x^2+x)\#}{\left(\frac{x^2+x}{2}\right)\#}\right) > \text{lcm}(x^2+x)$$

• if a prime $$p \ge (x+1)$$, then $$p^2 > (x^2+x)$$

• if a prime $$(x^2+x) \ge p^a > \frac{x^2+x}{2}$$, then $$p^{a+1} > (x^2+x)$$ and $$p^{a-1} \le \frac{x^2+x}{2}$$

(2) $$\text{lcm}(x^2+x) \ge {{x^2+x}\choose{\frac{x^2+x}{2}}}$$

This follows from Legendre's Formula since:

• Let $$v_p(x)$$ be the highest power of $$p$$ that divides $$x$$

• $$v_p({{x^2+x}\choose{\frac{x^2+x}{2}}}) = \sum\limits_{i \ge 1}\left\lfloor\dfrac{x^2+x}{p^i}\right\rfloor - 2\left\lfloor\dfrac{x^2+x}{2p^i}\right\rfloor$$

• It is well known that for each $$i$$, the difference is at most $$1$$ and that if $$i > \log_p(x^2+x)$$, then the difference is $$0$$.

(3) $${{x^2+x}\choose{\frac{x^2+x}{2}}} > \dfrac{2^{x^2+x}}{\frac{x^2+x}{2}}$$

For $$x \ge 4$$, $${{2x}\choose{x}} \ge \dfrac{4^x}{x}$$ since $${8\choose4} = 70 > \dfrac{4^4}{4} = 64$$ and $${{2x}\choose{x}} = 2\left(\dfrac{2x-1}{x}\right){2(x-1)\choose{x-1}} > 2\left(\dfrac{2x-1}{x}\right)\left(\dfrac{4^{x-1}}{x-1}\right) > \dfrac{4^x}{x}$$

(4) Using Hanson's result that $$\text{lcm}(x) < 3^x$$, gives:

$$\left(\frac{(x^2+x)\#}{\left(\frac{x^2+x}{2}\right)\#}\right) > \left(\frac{4}{3}\right)^{(x^2+x)/2}\left(\frac{2}{(x^2+x)3^x}\right)$$