There is also a b) part to this task, which I can't seem to solve

__It goes__:

Show that $\displaystyle a_{n}^2 -2 \geq 0$ for all values of n __This is what I'm thinking__:

If we assume that: $\displaystyle a_{n}^2 \geq 2$, we have to show that this is also the case for $\displaystyle a_{n+1}^2 \geq 2$

From before we have that: $\displaystyle a_1 = 2$ and $\displaystyle a_{n + 1} = \frac{a_n^2 + 2}{2a_n}$. And we also know that: $\displaystyle a_{n} \geq 0$ for all n.

I have tried putting $\displaystyle a_{n+1}^2 = (\frac{a_n^2 + 2}{2a_n})^2$.

But from here I'm not able to show that $\displaystyle a_{n+1}^2 \geq 2$

Any ideas?