If $ \ x_1=2 , \ x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n} \ $ ; show that $ \ x_n^2-2>0 \ $ for every index $ \ n \ $

Question

If $ \ x_1=2 , \ x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n} \ $ ; show that $ \ x_n^2-2>0 \ $ for every index $ \ n \ $.

Answer:

For $ \ n=1 \ $,

$ x_1^2-2=2^2-2=2>0 \ $

For $ \ n=2 \ $,

$ x_2^2-2=(\frac{x_1}{2}+\frac{1}{2})^2-2=(1.5)^2-2>0 \ $

Let for $ \ n=m \ $,

$ x_m^2-2>0 \ $

Now we have to show that $ \ x_{m+1}^2-2>0 \ $

But how to show that $ \ x_{m+1}^2-2>0 \ $ ?

Help me

Answer 1

Hint: prove $x_{n+1}^2-2=(\frac{x_n}{2}-\frac{1}{x_n})^2$.