This is only to add on the other answer:

Note that we have $m = m_1 + m_2$ and $-j_1 \le m_1 \le j_1$, so $$-j_1 \le m - m_2 \le j_1$$ In particular $m$ can take its maximum value $j$ and $m_2$ can take its maximum value $j_2$, which gives $$-j_1 \le j - j_2 \le j_1$$ Rearranging gives $$j_2-j_1 \le j \le j_1 + j_2$$

A similar argument with the indices $1$ and $2$ exchanged gives

$$j_1-j_2 \le j \le j_1 + j_2$$

This is only to add on the other answer:

Note that we have $m = m_1 + m_2$ and $-j_1 \le m_1 \le j_1$, so $$-j_1 \le m - m_2 \le j_1$$ In particular $m$ can take its maximum value $j$ and $m_2$ can take its maximum value $j_2$, which gives $$-j_1 \le j - j_2 \le j_1$$ In fact one can consider the situation where $m$ takes its minimum value $-j$ and $m_2$ takes its minimum value $-j_2$, which gives $$-j_1 \le -j + j_2 \le j_1$$ But this is in fact the same inequality after rearrangement.

The other two situations when one takes its minimum and the other takes its maximum are forbidden.