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

Write $$x_{n+1}^2 -2 =  \left(\frac{1}{2} \left(x_n+\frac{2}{x_n}\right)\right) ^2 -2 = \frac{x_n^2}{4}+\frac{1}{x_n^2} -1 =\left( \frac{x_n}{2} - \frac{1}{x_n}\right)^2 \ge 0  $$ as any number squared is positive. This shows that $x_n^2 \ge 2$ for all choices of $n$.
