We have the sharper inequality \[\left\vert \sqrt{2}a-b \right\vert \geq \frac{1}{\sqrt{2}a+b}.\] See that $|2a^{2}-b^{2}| \geq 1 = \frac{|\sqrt{2}a+b|}{\sqrt{2}a+b}$ and divide through $|\sqrt{2}a+b|$, voila.

Peter wrote: Prove that for any positive integers $ a$ and $ b$ \[ \left\vert a\sqrt{2}-b\right\vert >\frac{1}{2(a+b)}.\] Reference for this problem [GML, pp. 183] is wrong. The correct reference is (A. Mirotin) Belarus 2002.

ZetaX very nice solution