Diophantine bound for the square root of two ($\Vert q\sqrt{2}\Vert \ge 1/(4q)$)

For the nearest integer $a$ to $q\sqrt{2}$, the nonzero integer $2q^2-a^2$ has absolute value at least one. Dividing by $q\sqrt{2}+a\le 4q$ proves the bound. Thus the points $0,\sqrt{2},\dots ,W\sqrt{2}$ modulo one are separated by at least $1/(4W)$. Ordering their distances from zero gives ${\sum }_{h\le W}\Vert h\sqrt{2}{\Vert }^{-1}\ll W\log (2W)$.