## An inequality involving sqrt(2)

Prove that for all $a,b \in \mathbb{Z}^+$, $| \sqrt{2} - \frac{a}{b} | \geq \frac{1}{3b^2}$.

Let $p(x) = x^2 - 2$.

First we will show that the inequality holds for $b = 1$. Note that $|\sqrt{2} - \frac{1}{1}| = 0.414 + \epsilon > 1/3$ and $|\sqrt{2} - \frac{2}{1}| = 0.585 + \epsilon > 1/3$. For $a > 2$, we have $|\sqrt{2} - \frac{a}{1}| > 1 > 1/3$. So the inequality holds for $b = 1$. Henceforth, we will assume that $b \geq 2$.

Suppose $|\sqrt{2} - \frac{a}{b}| \geq \frac{3 - 2\sqrt{2}}{2}$. Note that since $b \geq 2$, $b^2 \geq 4 > 3.88 + \epsilon = \frac{2}{9 - 6\sqrt{2}}$; so $\frac{3 - 2\sqrt{2}}{2} > \frac{1}{3b^2}$. Hence $|\sqrt{2} - \frac{a}{b}| \geq \frac{1}{3b^2}$.

Now suppose $|\sqrt{2} - \frac{a}{b}| \leq \frac{3 - 2\sqrt{2}}{2}$. By the Mean Value Theorem from calculus (which we will assume to be valid), there exists an element $\xi$ between $\sqrt{2}$ and $a/b$ such that $p^\prime(\xi) = \frac{p(\sqrt{2}) - p(a/b)}{\sqrt{2} - a/b}$, and hence $|p^\prime(\xi)| = \frac{|p(\sqrt{2}) - p(a/b)|}{|\sqrt{2} - a/b|}$. Now $\xi \in [\sqrt{2} - \frac{3-2\sqrt{2}}{2}, \sqrt{2} + \frac{3-2\sqrt{2}}{2}] = [1.328+\epsilon, 1.5]$. Since $p^\prime(x) = 2x$ is strictly increasing, we have $|p^\prime(\xi)| \leq 3$. Since $p(\sqrt{2}) = 0$, we have $|p(a/b)| \leq 3|\sqrt{2} - \frac{a}{b}|$.

Note that $p(a/b) \neq 0$, since (for example) $p(x)$ is irreducible over $\mathbb{Q}$. Now $|p(a/b)| = |\frac{a^2}{b^2} - 2|$ $= \frac{|a^2 - 2b^2|}{b^2}$. Since $p(a/b) \neq 0$, $a^2 - 2b^2$ is a nonzero integer. In particular, we have $|a^2 - 2b^2| \geq 1$, so that $|p(a/b)| \geq \frac{1}{b^2}$.

Hence $|\sqrt{2} - \frac{a}{b}| \geq \frac{1}{3b^2}$.