Context (Though I don't see how this might help deriving the final inequality):
Example : For $p$ prime, $x$ greater than $0$ a real number, $n$ greater or equal to $2$ an integer: for $p< 2^{x}$ it holds that $1+\text{ord}_{p}(n)\le 1+ \frac{\log n}{\log p} \le 1+ \frac{\log n}{\log 2} \le \frac{2}{\log 2}\log n$ and for $p \ge 2^{x}$ it holds that: $1+\text{ord}_{p}(n)\le 2^{\text{ord}_{p}(n)}\le p^{\text{ord}_{p}(n)/x}.$
(no proof supplied from the original author, so I will try to do one myself.)
For $p<2^{x}$:
$1+\text{ord}_{p}(n)\le 1+ \frac{\log n}{\log p} \tag{1}$
$1+ \frac{\log n}{\log p} \le 1+ \frac{\log n}{\log 2} \tag{2}$
$1+ \frac{\log n}{\log 2} \le \frac{2}{\log 2}\log n \tag{3}$
(2) holds because $2$ is the smallest prime. (3) because the derivative of $\log(2x)$ is $\frac{1}{x}$ and $\frac{2}{x}$ (the derivative of $2\log(x)$) is greater for every $x$. I have trouble showing (1) because I don't understand how to handle the $\text{ord}_{p}(n)$ (which also appears in the second inequality). ($\text{ord}_{p}(n)$ is the highest natural exponent $k$ such that $p^{k}$ divides $n$).
For $p>2^{x}$: $1+\text{ord}_{p}(n)\le 2^{\text{ord}_{p}(n)} \tag{1}$
$2^{\text{ord}_{p}(n)}\le p^{\text{ord}_{p}(n)/x} \tag{2}$
Right after this example, the author mentions an inequality and this is what I am really interested in:
For $n\ge 2$ an integer, $ x\in \mathbb{R_{>0}}$ , how does one find out that $\tau (n) < \left(\frac{2}{\log 2}\log n \right)^{2^x}n^{\frac{1}{x}}\ ?$ It looks like a popular inequality, does it have a name?