Am I correct that this is established if I can show that:
- It is true for $x \ge 5$
- $f(x)=\dfrac{(2x-2)\ln x}{2.25506x}$ is increasing for $x \ge 5$.
If this assumption is correct, then the answer is yes since:
It is true for $x = 5$ since:
$$2(5)-2)\ln(5) > 12.8 > 11.3 > 2.25506\times5$$
It is increasing for $x \ge 1$ since:
(1) Using the quotient rule with $g(x) = (2x-2)\ln n$ and $h(x)=2.25506x$:
$$f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$$
(2) Using the product rule with $s(x) = (2x - 2)$ and $t(x) = \ln(x)$
$$g'(x) = s'(x)t(x) + s(x)t'(x) = 2\ln(x) + \dfrac{2x - 2}{x} = \dfrac{2x\ln(x) + 2x - 2}{x}$$
(3) So that:
$$f'(x) = \frac{(2x\ln(x) + 2x - 2)2.25506 - (2x-2)\ln(x)(2.25506) }{(2.25506x)^2} = \frac{2x\ln(x) + 2x - 2 - (2x-2)\ln(x)}{2.25506x^2} = \frac{2x - 2 +2\ln(x)}{2.25506x^2}$$
(4) $\frac{2x - 2 +2\ln(x)}{2.25506x^2} > 0$ for $x \ge 1$
Edit: Change $n$ to $x$ in the second bullet based on a comment.
Much easier is to note $ \ln 5 \gt 1.6$ so $$(2x-2) \ln x \gt 3.2x-3.2 =2.22506x +(0.97494x-3.2) \gt 2.22506x$$ when $x \gt 5$
Your approach of showing that the ratio starts out greater than $1$ and is increasing is fine, too.