Is $n \log n$ in $O(n^{1.46-\varepsilon})$?
When $f(n)$ is said to be polynomially smaller than $g(n)$ it just means that there is some constant $\varepsilon &gt; 0$ such that $f(n) \in O(g(n) n^{-\varepsilon})$.
This is not about being able to bound $f(n)$ from above and/or below but about the growth rate of $f(n)$ when compared to $g(n)$.
In other words you want $\frac{g(n)}{f(n)}$ to  grow at least as fast as some polynomial/root of $n$.
In your particular case, $n \log n$ is indeed polynomially smaller than $n^{1.46}$, and this follow from the fact that $n \log n \in O(n^\alpha)$ for any $\alpha &gt;1$.
A a concrete choice of $\varepsilon$, you can pick $\varepsilon=0.1$ yielding $n \log n \in O(n^{1.46-0.1}) = O(n^{1.36})$.
To see this you can consider the limit:
$$
\lim_{n \to \infty} \frac{n \log n}{n^{1.36}} =
\lim_{n \to \infty} \frac{\log n}{n^{0.36}}  =
\lim_{n \to \infty} \frac{1}{0.36 \cdot n^{0.36}} = 0,
$$
where we used L'Hôpital's rule.
In the above, the choice $\varepsilon = 0.1$ was just a convenient value that works, but any choice of $\varepsilon \in (0, 0.46)$ suffices.