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 (a) you don't need to prove the statement for $a_2$.
You can simplify the argument by proving first that $a_n\ge0$. This is true for $a_1$; suppose it is true for $a_n$; then $a_{n+1}=\sqrt[3]{(a_n+1)/2}\ge0$.
Now the other inequality, which you did well: if $a_n\le1$, then $(a_n+1)/2\le1$ and therefore $a_{n+1}=\sqrt[3]{(a_n+1)/2}\le1$.
For (b) the arrows are in the wrong direction! You need to prove that $a_{n+1}\ge a_n$ and you can't take this as an assumption. However
$$
a_{n+1}\ge a_n
\quad\text{if and only if}\quad
a_{n+1}^3\ge a_n^3
\quad\text{if and only if}\quad
\frac{a_{n}+1}{2}\ge a_n^3
$$
so we are reduced to proving that $2a_n^3\le a_n+1$.
Your argument is fine: $a_n\le1$ implies $a_n^3\le a_n$ and $a_n^3\le 1$.
