Prove that if $-2 \leq x_0 \leq 2$, then $-2 \leq 3x_0 - x_0^3 \leq 2$.

Question

Given $-2 \leq x_0 \leq 2$, I have to prove $-2 \leq x_n \leq 2$, where $x_n = 3x_{n - 1} - x_{n - 1}^3$ and $n \in \mathbb{Z}_+$. I thought this problem can be solve by mathematical induction, but I can't prove the initial step, i.e., $n = 1$.

I try this two ways:

1st. If $-2 \leq x_0 \leq 2$, then $-6 \leq 3x_0 \leq 6$ and $-8 \leq x_0^3 \leq 8$. But, this provied the inequality

\begin{equation} -14 \leq 3x_0 - x_0^3 \leq 14, \end{equation}

that its true.

2nd. I try prove it separately, with $0 \leq x_0 \leq 2$ first, but its wrong too.

How can I prove it?

Answer 1

There is an elementary way to prove this using AM-GM. If $x^2<3$, then

$$ \left( x(3-x^2) \right)^2 =\frac{1}{2} 2x^2 \cdot (3-x^2) \cdot (3-x^2)\\ \leqslant \frac{1}{2} \left(\frac{2x^2+3-x^2+3-x^2}{3} \right)^3 =4. $$

If $x^2\ge 3$, then $$ \left( x(3-x^2) \right)^2 =x^2 \cdot (x^2-3)^2 \leqslant 4\cdot 1^2=4. $$

Answer 2

$$f(x)=3x-x^3 \implies f'(x)=3-3x^2, f'(x)=0 \implies x=0,\pm 1$$ So $$f_{max}=f(1)=2, f(2)=-2, f(0)=0 \implies -2\le f(x) \le 2$$

Answer 3

Prove that $f(x)+2$ is positive, and $f(x)-2$ is negative. Hint: factors