Approximating $x=\sqrt{2}+1$

Let$ y<1+\sqrt{2}$,

So $1+\sqrt{2} - y=D$ (say).

$1/y > \sqrt{2}-1$

And $1/y+2>1+\sqrt{2}$

So let $1/y+2-(1+\sqrt{2}) =d.$

So we have to prove d < D.

Let D-d>0.

Simplifying you get : 2$\sqrt{2} >y+1/y $

Which is true as $1-\sqrt{2}<y<1+\sqrt{2}$.

You can do similarly for $y>1+ \sqrt{2}$