In Waclaw Sierpinski's book Elementary Theory of Numbers on page 168 there is the following exercise:
"Exercises. 1. Prove that for natural numbers $n$ we have $d(n) \leq 2\sqrt{n}$," where $d(n)$ is the number of divisors of n.
As a hint right below is given: "The proof follows from the fact that of two complementary divisors of a natural number $n$ one is always not greater than $\sqrt{n}$.
I understand the hint but I don't know how it can be used to prove $d(n)\leq 2\sqrt{n}$.
Complementary divisors are pairs of divisors that when multiplied gives the number that is to be divided. For example the number $120$ has the complementary divisors: \begin{align*} & 1, 120 \\ & 2, 60 \\ & 3, 40 \\ & 4, 30 \\ & 5, 24 \\ & 6, 20 \\ & 8, 15 \\ & 10, 12 \\ \end{align*} How do you prove that $d(n) \leq 2\sqrt{n}$?