Show that $\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$ for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$. Show that $$\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$$
for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$.
I tried many different ways to expand left side and estimate it but always got stuck at some point.
 A: As in How find the value $\beta$ such $\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^2}$ (and the above comments),
$$
\left| \sqrt2-\frac{h}{k} \right| \, \left| \sqrt2+\frac{h}{k} \right|
= \left| 2-\frac{h^2}{k^2} \right| = \frac{|2k^2 - h^2|}{k^2} \ge
 \frac{1}{k^2} \tag{1}
$$
for all  $k \in \mathbb{N}$ and $h \in \mathbb{Z}$, because $\sqrt 2$ is irrational.
Now assume that there exist some $k \in \mathbb{N}$ and $h \in \mathbb{Z}$ such that
$$
\left| \sqrt2-\frac{h}{k} \right| < \frac {1}{4k^2} \quad . \tag 2
$$
Then the other factor can be estimated as
$$
\left| \sqrt2+\frac{h}{k} \right| = \left| 2 \sqrt 2 - \bigl(\sqrt 2 - \frac{h}{k} \bigr) \right| \le
 2 \sqrt 2 + \left| \sqrt2-\frac{h}{k} \right| \\
 <  2 \sqrt 2 + \frac {1}{4k^2} \le  2 \sqrt 2 + \frac 14\approx 3.078 < 4 \quad . \tag 3
$$
Multiplying the inequalities $(2)$ and $(3)$ gives a contradiction 
to $(1)$. Therefore $(2)$ must be false 
for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$,
which is what you wanted to prove.
A: I will show that
$|\sqrt{n}-\frac{x}{y}|
>\frac1{(\sqrt{n}+\sqrt{n+1})y^2}
$.
For $n=2$,
$\sqrt{2}+\sqrt{3}
< 3.15
$,
so
$|\sqrt{2}-\frac{x}{y}|
>\frac1{3.15 y^2}
$.
I also show that if
$|\sqrt{n}-\frac{x}{y}|
<\frac1{(2\sqrt{n}+\epsilon)y^2}
$.
then
$y
<\sqrt{ \frac1{2\epsilon\sqrt{n}}}
$.
In general,
if $n$ is not a perfect square,
then
$|x^2-ny^2| \ge 1
$
for integers $x$ and $y$.
Therefore,
if $z = \sqrt{n}$,
$\begin{align*}
1
&\le |x^2-ny^2|\\
&=|x-zy||x+zy|\\
\text{so}\\
|x-zy|
&\ge \frac1{|x+zy|}\\
&= \frac1{x+zy}\\
\text{or}\\
|z-\frac{x}{y}|
&\ge \frac1{y}\frac1{x+zy}\\
&\ge \frac1{y^2}\frac1{z+x/y}\\
\end{align*}
$
If
$|z-\frac{x}{y}|
< \frac1{cy^2}
$,
then
$\frac{x}{y}
< z+\frac1{cy^2}
$
and
$\frac1{z+x/y}
\le y^2|z-\frac{x}{y}|
<\frac1{c}
$
so that
$c 
<z+x/y
<z+z+\frac1{cy^2}
=2z+\frac1{cy^2}
$.
Therefore
$c^2-2zc
< \frac1{y^2}
\le 1
$
or
$c^2-2zc+z^2
< z^2+1
$
or
$(c-z)^2
< n+1
$
or
$c < z+\sqrt{n+1}
=\sqrt{n}+\sqrt{n+1}
$.
Therefore
$|\sqrt{n}-\frac{x}{y}|
>\frac1{(\sqrt{n}+\sqrt{n+1})y^2}
$.
For $n=2$,
$c <
\sqrt{2}+\sqrt{3}
< 3.15
$.
(added later)
Note that,
since
$c^2-2zc
< \frac1{y^2}
$,
$c < z+\sqrt{z^2+\frac1{y^2}}
=z+z\sqrt{1+\frac1{z^2y^2}}
<z+z(1+\frac1{2z^2y^2})
=2z+\frac1{2zy^2})
$.
Therefore,
if $c > 2z$,
$\frac1{2zy^2}
> c-2z
$
or
$2zy^2
< \frac1{c-2z}
$
or
$y
<\sqrt{ \frac1{2z(c-2z)}}
$.
