Can this be proved without making use of derivatives?

Question

Problem:

Let $\left(a_{n}\right)$ be a sequence with $a_{1}=1$ and $a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{2}{a_{n}}\right)$.

It must be proved that $a_{n}\geq\sqrt{2}$ for $n\geq2$.

I have a proof, but my professor wants a proof that is not using differentation.

Question1: can this be proved purely on base of induction and without making use of differentiating?

Question2: is my proof okay?

My proof:

function $f\left(x\right)=\frac{1}{2}\left(x+\frac{2}{x}\right)$ has $f'\left(x\right)=\frac{1}{2}-x^{-2}$ as derivative so that $f'\left(\sqrt{2}\right)=0$. We have $f'\left(x\right)>0$ if $x>\sqrt{2}$ and $f'\left(x\right)<0$ if $0<x<\sqrt{2}$. Conclusion: there is minimum at $x=\sqrt{2}$ that takes value $f\left(\sqrt{2}\right)=\sqrt{2}$ and $f$ is increasing on $\left(\sqrt{2},\infty\right)$.

The last fact is enough to prove by induction that $a_{n}>\sqrt{2}$ for $n\geq2$:

$a_{2}=1.5>\sqrt{2}$ and $a_{n+1}=f\left(a_{n}\right)>f\left(\sqrt{2}\right)=\sqrt{2}$ based on hypothese $a_{n}>\sqrt{2}$.

Thanks in advance.

Answer 1

First note that $a_n \geq 0$ for all $n \in \mathbb{N}$, so you can use AM-GM inequality:

$$\frac{a_n+\frac{2}{a_n}}{2} \geq \sqrt{a_n \cdot \frac{2}{a_n}}=\sqrt{2}$$.

Answer 2

Or if AM-GM doesn't strike your mind at the moment then you can complete the square like this $$a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{2}{a_{n}}\right)=\frac{1}{2}\left(\sqrt{a_{n}}-\sqrt{\frac{2}{a_{n}}}\right)^2+\sqrt{2}\ge \sqrt{2}$$