Proving $\lfloor na_n \rfloor = 0$ if $a_n > 0$ and $a_n^2<a_n-a_{n+1}$

Question

If $a_n > 0$ and $a_n^2<a_n-a_{n+1}$ for any positive integer $n$, prove that:

$$\lfloor na_n \rfloor = 0$$

So $na_n > 0$ and I tried to prove that $a_n < \frac{1}{n}$ with induction.

Base step $n = 1$: $a_1^2 < a_1 - a_2 < a_1 \implies a_1<1=\frac{1}{1}$

Induction step $a_n < \frac{1}{n}$: $a_{n+1}<a_n-a_n^2$ and here I am stuck because I can only prove $a_n-a_n^2 < \frac{1}{4}$ here.

Answer 1

Hint: The function $f(x) = x-x^2$ is increasing on $[0,\tfrac{1}{2}]$. So if $0 < a_n < \tfrac{1}{n}$, then $f(0) < f(a_n) < f\left(\tfrac{1}{n}\right)$.

Can you take it from here?