Distance to nearest integer inequality

Question

If $\langle a \rangle$ denotes the distance from the real number $a$ to the nearest integer then why is it that for $a,b \in \mathbb{R}$ we have:

$$min \langle a \pm b \rangle \leqslant | \langle a \rangle - \langle b \rangle |$$

Answer 1

The result you want is actually $\min \langle a \pm b \rangle \leqslant | \langle a \rangle - \langle b \rangle |$ not that both inequalities are true as that clearly implies the required inequality in the OP.

Proof: $a=m+c, b=n+d, |c| \le 1/2, |d| \le 1/2, m, n \in \mathbb Z$ then clearly $\min |c\pm d| \le 1/2$ as one can take the minus sign if $cd \ge 0$ and the plus sign if $cd \le 0$ and get the required inequality, hence $\min \langle a \pm b \rangle = \min |c \pm d|$.

But now $| \langle a \rangle - \langle b \rangle| = |c-d|$ so if $cd \ge 0$ we get equality in the required inequality as $\min |c \pm d|=|c-d|$ while if $cd <0$ then clearly $|c-d| >|c+d|=\min |c \pm d|$ (easily seen by squaring for example), so we get strict inequality. In any case, the required inequality, hence the book result is proven.