Where did I go wrong in my proof that for all $n \in \mathbb{Z}^+$, $\sqrt{2} < a_n$ with $(a_n)$ being a particular recursive sequence? I am in Introduction to Abstract Math, and I have taken Calculus 1, Linear Algebra, and Discrete Math. I got stuck with an apparent contradiction in my proof and wanted to know where I messed up:
Consider the sequence $(a_n)$ defined recursively by:
$$a_1 = 3, \:\:\:\:\:\:\: a_{n+1} = \frac{a_n}{2}+\frac{1}{a_n} \:\:\:\:\:\:\: (n \geq 1).$$
Prove that $\sqrt{2}$ is a lower bound for $(a_n)$.
This can be rewritten as: Prove for all $n \in \mathbb{Z}^+$, $\sqrt{2} < a_n$, which we can prove with strong induction.
Base Step
Check for $n=1$.
$$a_1 = 3 > \sqrt{2}$$
Check for $n=2$.
$$a_2 = \frac{a_1}{2} + \frac{1}{a_1} = \frac{3}{2} + \frac{1}{3} = \frac{11}{6} > \sqrt{2}$$.
Therefore, the statement holds true for $n=1, n=2$.
Inductive Step
Assume that $a_n > \sqrt{2}$ for $1 \leq n \leq k$ with $k \in \mathbb{Z}^+$. Prove $a_{k+1} > \sqrt{2}$.
$$a_k = \frac{a_{k-1}}{2} + \frac{1}{a_{k-1}} > \sqrt{2}$$
$$a_{k+1} = \frac{a_k}{2} + \frac{1}{a_k}$$
$$a_{k+1} = \frac{\frac{a_{k-1}}{2} + \frac{1}{a_{k-1}}}{2} + \frac{1}{\frac{a_{k-1}}{2} + \frac{1}{a_{k-1}}}$$
$$a_{k+1} = \frac{2a_{k-1}}{a_{k-1}^2 + 2}+\frac{a_{k-1}}{4}+\frac{1}{2a_{k-1}}$$
$$a_{k+1} = \frac{2a_{k-1}}{a_{k-1}^2 + 2}+\frac{1}{2}(\frac{a_{k-1}}{2}+\frac{1}{a_{k-1}})$$
$$a_{k+1} = \frac{2a_{k-1}}{a_{k-1}^2 + 2}+\frac{1}{2}(a_k)$$
$$a_{k+1} = (\frac{2a_{k-1}}{a_{k-1}^2 + 2} \times \frac{1}{a_k})a_k+\frac{1}{2}(a_k)$$
$$a_{k+1} = a_k (\frac{2a_{k-1}}{a_{k-1}^2 + 2} \times (\frac{2}{a_{k-1}}+a_{k-1}) +\frac{1}{2})$$
Here's where I got stuck. I plugged it into Wolfram Alpha to do the algebra for that step and see if there was equivalent forms, and I saw that what was in the parenthesis was equal to $\frac{5}{2}$, which gives
$$a_{k+1} = \frac{5a_{k}}{2}$$
which doesn't make sense as the sequence would be increasing way too fast.
Where did I mess up in my proof?

As another approach, let $f(x)=\frac x2+\frac 1x$ for $x>0$.  It is easy to confirm that $$f(x)=\sqrt 2\iff x=\sqrt 2$$
As $f(x)$ is clearly continuous for all $x>0$ we see that $x>\sqrt 2$ must then imply that $f(x)>\sqrt 2$ and we are done.
