Proving $\left|\sqrt2-(a/b)\right|\geq1/(3b^2)$ This is Problem 4.26 on p.58 of The Theory of Algebraic Numbers by Harry Pollard and Harold G. Diamond (Dover edition).

Prove that $\left|\sqrt2-\dfrac ab\right|\geq\dfrac1{3b^2}$ for all positive integers $a,b$.

Here's what I've done so far. Split into cases:


*

*(i) $\dfrac ab<\sqrt2$

*(ii) $\dfrac ab>\sqrt2+\dfrac13$

*(iii) $\sqrt2<\dfrac ab<\sqrt2+\dfrac13$.


(i) I followed the method used by the authors in proving a similar result in that chapter. Let $f(x)=x^2-2$; this is the minimum polynomial of $\sqrt2$ in $\mathbb Q$ so $f\left(\sqrt2\right)=0$. By the mean-value theorem:
$$\frac{f\left(\sqrt2\right)-f\left(\dfrac ab\right)}{\sqrt2-\dfrac ab}\ =\ f'(\xi)$$
for some $\dfrac ab<\xi<\sqrt2$. We have $f'(\xi)=2\xi<2\sqrt2<3$, and so
$$\left|f\left(\frac ab\right)\right|\ <\ 3\left|\sqrt2-\frac ab\right|$$
But
$$\left|f\left(\frac ab\right)\right|=\left|\frac{a^2}{b^2}-2\right|=\frac{|a^2-2b^2|}{b^2}\geq\frac1{b^2}$$
Hence
$$\left|\sqrt2-\frac ab\right|\ \geq\ \frac1{3b^2}$$
as required.
(ii) Done immediately as
$$\left|\sqrt2-\frac ab\right|=\frac ab-\sqrt2\,> \frac13\geq\frac1{3b^2}$$
(iii) This is where I'm stuck.
So I've basically done much of the hard work above and the final piece is all I need to complete the jigsaw. I would be grateful for some help. Thanks.
 A: Do the case distinction with $1/12$ instead of $1/3$. This does not affect (i) and (iii) works in the same way as (i),  since you still have $2 \xi < 3$ as $2(\sqrt{2} +1/12)<3$. 
Of course, you need to reconsider (ii) then. But it is hardly more difficult, as you have $\frac{a}{b} - \sqrt{2} \ge \frac{1}{12}$ and 
$\frac{1}{12} = \frac{1}{3 \cdot 4^2} \ge \frac{1}{3b^2}$, except if $b=1$.
Yet then if $b=1$, you have that $a/b$ is an integer so that in this case $a/b \ge 2$ and the difference is at least $1/3$ in this case too.   
There may be a shorter method, but this was what came to mind.
A: 
Prove that $$\left|\sqrt2-\dfrac ab\right|>\dfrac1{3b^2}$$ for integers $a$ and $b$ with $b\neq 0$.

Without loss of generality, assume that $b>0$.  Note that
$$\left|\sqrt{2}-\frac{a}{b}\right|=\frac{\left|a^2-2b^2\right|}{b(a+\sqrt{2}b)}\,.\tag{*}$$
If $a<0$, then
$$\left|\sqrt{2}-\frac{a}{b}\right|>\sqrt{2}>\frac{1}{3b^2}\,.$$
If $0\leq a\leq b$, then, using (*),
$$\left|\sqrt{2}-\frac{a}{b}\right|\geq\frac{b^2}{b(b+\sqrt{2}b)}\geq \frac{1}{1+\sqrt{2}}>\frac{1}{3}\geq\frac{1}{3b^2}\,.$$
If $a\geq \frac{3}{2}b$ and $b>1$, then
$$\left|\sqrt{2}-\frac{a}{b}\right|\geq\frac{3}{2}-\sqrt{2}>\frac{1}{12}\geq\frac{1}{{3b^2}}\,.$$
If $a>1$ and $b=1$, then
$$\left|\sqrt{2}-\frac{a}{b}\right|\geq 2-\sqrt{2}>\frac{1}{3}\geq\frac{1}{3b^2}\,.$$
If $b<a<\frac{3}{2}b$, then, using (*),
$$\left|\sqrt{2}-\frac{a}{b}\right|\geq \frac{1}{b\left(a+\sqrt{2}b\right)}>\frac{1}{b\left(\frac{3}{2}b+\sqrt{2}b\right)}=\frac{1}{\left(\frac{3}{2}+\sqrt{2}\right)b^2}>\frac{1}{3b^2}\,.$$
Indeed, the smallest constant $\gamma>0$ such that $$\left|\sqrt{2}-\frac{a}{b}\right|\geq \frac{1}{\gamma\,b^2}$$ for all integers $a$ and $b$ with $b\neq 0$ is $\gamma=\frac{3}{2}+\sqrt{2}$.  The equality holds if and only if $a=3$ and $b=2$.

Related Observation

Conjecture.  Let $\gamma_b:=\max\,\left\{\frac{1}{b^2\,\left|\sqrt{2}-\frac{a}{b}\right|}\,\Big|\,a\in\mathbb{Z}\right\}$ for each $b\in\mathbb{N}$.  Define $$\Gamma:=\liminf_{b\to\infty}\,b\cdot\gamma_b\,.$$  Then, $$\Gamma=\inf\,\left\{b\cdot\gamma_b\,\big|\,b\in\mathbb{N}\right\}=2\,.$$

After checking all positive integers $b\leq 1500$, I find that $$\Gamma\leq 1+\frac{29}{41}\sqrt{2}\lesssim 2.0003\,.$$  Below is a plot illustrating the credibility of this conjecture.  Th horizontal axis is $b$.  The fuzzy blue line shows $\gamma_b$ versus $b$, and the nice orange line is $\dfrac{2}{b}$ versus $b$ for $b=1,2,\ldots,1500$.

What is very strange is $\Gamma$ seems to be equal to $2$ as well if $\sqrt{2}$ is replaced by $\sqrt{3}$, $\sqrt{5}$, $\sqrt{6}$, $\sqrt{7}$, and $\sqrt{8}$.  Could $\Gamma=2$ universally hold when $\sqrt{d}$ replaces $\sqrt{2}$ for any non-square positive integer $d$?  I tried replacing $\sqrt{2}$ by $\frac{1+\sqrt{5}}{2}$, $\sqrt[3]{2}$, $\pi$, $\text{e}$, and $\ln(3)$ as well, and it looks like $\Gamma=2$ still holds.
On the other hand, $$\limsup\limits_{b\to\infty}\,\gamma_b=2\sqrt{2}\,.$$  If $\sqrt{2}$ is replaced by $\sqrt{d}$ for any non-square positive integer $d$, then $$\limsup\limits_{b\to\infty}\,\gamma_b=2\sqrt{d}\,.$$
A: Here is the alternate way, without case analysis.
Suppose that for some $a,b$, $|\sqrt 2 - {a \over b}| < {1 \over kb^2}$. Rewriting the inequality,
$$-{1 \over kb^2} < {a \over b} - \sqrt 2 < {1 \over kb^2}\\
2-{2\sqrt 2 \over kb^2} + {1 \over k^2b^4} < {a^2 \over b^2} < 2 + {2\sqrt 2 \over kb^2} + {1 \over k^2b^4} \\
a^2 = 2b^2 + {1 \over k^2b^2} + {\delta \over k}$$
where $|\delta| < 2 \sqrt 2$. Also, from $\sqrt 2$ being irrational we have
$$|a^2 - 2b^2| \ge 1 \\
|{1 \over k^2b^2} + {\delta \over k}| \ge 1 \\
|\delta| \ge k - {1 \over kb^2} \\
k - {1 \over kb^2} < 2 \sqrt2 \\
k < \sqrt2  + \sqrt{2 + {1 \over b^2}}$$
Finally for $k=3$ $b=1$ can be excluded separately and $\sqrt 2 + \sqrt{2 + {1 \over 4}} < 3.$
