Let $a_{n+1}=\sqrt[3]{\frac{1+a_{n}}{2}}$ show that $0 \leq a_{n} \leq 1$ Let $0 \leq a_{1} \leq 1$ and $a_{n+1}=\sqrt[3]{\frac{1+a_{n}}{2}}$ for all $n \geq 1$.
Show that:
(a) $0 \leq a_{n} \leq 1$ for all n.
(b) $ a_{n+1} \geq a_{n}$ for all n.
I don't know if the following is correct.
My attempt.
By using induction:
(a) Show that $0 \leq a_{n} \leq 1$ for all n.
We have $0 \leq a_{1} \leq 1$ and we want to show that $0 \leq a_{2} \leq 1$ for all n.
$0 \leq a_{1} \leq 1 \Rightarrow \frac{1}{2} \leq \frac{a_{1}+1}{2} \leq 1 \Rightarrow  \sqrt[3]{\frac{1}{2}} \leq  \sqrt[3]{\frac{a_{1}+1}{2}}  \leq \sqrt[3]1 \Rightarrow 0.8 \leq a_{2} \leq 1 \Rightarrow 0 \leq a_{2} \leq 1 $
Now we suppose that the relation is true for n.
We want to show that it is also true for n+1.
We have:
$0 \leq a_{n} \leq 1 \Rightarrow \frac{1}{2} \leq \frac{a_{n}+1}{2} \leq 1 \Rightarrow  \sqrt[3]{\frac{1}{2}} \leq  \sqrt[3]{\frac{a_{n}+1}{2}}  \leq \sqrt[3]1 \Rightarrow 0.8 \leq a_{n+1} \leq 1 \Rightarrow 0 \leq a_{n+1} \leq 1 $
(b) Show that $ a_{n+1} \geq a_{n}$ for all n.
$ a_{n+1} \geq a_{n} \Rightarrow \sqrt[3]{\frac{1+a_{n}}{2}} \geq a_{n} \Rightarrow 1+a_{n} \geq 2a_{n}^3$
This is true because we know that $0 \leq a_{n} \leq 1$ and that $ a_{n} \geq a_{n}^3$.

For $(b)$ it doesnt look like you've shown the base case nor the implication that if $a_{n+1} \geq a_{n}$, then $a_{n+2} \geq a_{n+1}$. All you did was reach a true statement...
I see what you have done now, but it is less obvious that you are meaning to say that since these steps are reversible, the inequality is satisfied.
I would still write out that you have reversible steps, or even start your implications starting at $1+a_{n} \geq 2{{a_n}^3}$ and showing this is true since by the inequalities you have listed we have $1+ a_{n} \geq a_{n}+a_{n}=2a_{n} \geq 2a^3_{n}$. It looks like your way might be a bit better because it follows directly from proving the base case in my proof.
I did it a different way:
Suppose $a_{n+1} \geq a_{n}$ is true for some $n$ (it is clear from the previous work this works for $n=1$).
Then $a_{n+2} = \sqrt[3]{\frac{1+a_{n+1}}{2}} \geq? \sqrt[3]{\frac{1+a_{n}}{2}} = a_{n+1}$ which is true by the inductive hypothesis.
In fact, the base case for $(b)$ might not be so clear without using your inequality anyways.
