Let $a_{n+1}=a_n-a_n^3$ with $a_1=\frac{1}{3}$. Show that $a_n>0$ for all $n\in\mathbb{N}$

Question

I come across an issue when I try to solve it with induction. Namely, I assume as induction hypothesis that $a_n>0$ for a certain $n$ and proceed to consider $a_{n+1}>0$ this gives $a_n-a_n^3>0$. But I run into an issue, namely take $a_n=2<0$ but this certainly gives $a_{n+1}=2-8=-6<0$. So the assumption $a_n>0$ doesn't get help me anyhow. Do you have any hints as to what my induction hypothesis should be?

Answer 1

Hint: $a_n<1$ for all $n$.

Answer 2

It is easy to see that $a_1,a_2,a_3$ are positive. Let $n>3$ and suppose $a_{n+1}$ is first negative term. Then $$a_n(1-a_n)(1+a_n)<0 \implies a_n>1$$

let $x=a_{n-1}$ so $$x>1+x^3 = (x+1)(x^2-x+1)\geq (x+1)x \implies x^2<0$$ a contradiction.