Show that $\left(\frac{1}{2}\left(x+\frac{2}{x}\right)\right)^2 > 2$ if $x^2 > 2$ Okay, I'm really sick and tired of this problem. Have been at it for an hour now and we all know the drill: if you don't get to the solution of a simple problem, you won't, so ...
I'm working on a proof for the convergence of the Babylonian method for computing square roots. As a warming up I'm first using the sequence $(x_n)$ defined by:
$$
x_1 = 2\\
x_{n+1} = \frac{1}{2} (x_n + \frac{2}{x_n})
$$
Now for the proof, I want to show that: $\forall n \in \mathbb{N}: x^2_n > 2$. I want to prove this using induction, so this eventually comes down to:
$$
x_n^2 > 2 \implies x_{n+1}^2 = \frac{1}{4}x_n^2 + 1 + \frac{1}{x_n^2} > 2
$$
And I can't seem to get to the solution. Note that I don't want to make use of showing that $x=2$ is a minimum for this function using derivatives. I purely want to work with the inequalities provided. I'm probably seeing through something very obvious, so I would like to ask if anyone here sees what's the catch.
Sincerely,
Eric

Just in case you don't insist on using induction.
$$\left(x+\frac2x\right)^2 \ge 8 \Leftrightarrow x^2+4+\frac4{x^2} \ge 8 \Leftrightarrow x^2-4+\frac4{x^2}\ge 0 \Leftrightarrow \left(x-\frac2x\right)^2\ge 0$$
The equality holds if and only if $x-\frac2x=0$, i.e. if $x^2=2$.
This is very similar to the usual derivation of AM-GM inequality for two variables.
(Or you could use directly AM-GM, if you are familiar with it.)
