Prove complex number inequality

Question

$$
\left| {\frac{{z_1 + z_2 }}{{z_3 + z_4 }}} \right| \le \frac{{\left| {z_1 } \right| + \left| {z_2 } \right|}}{{\Big | {\left| {z_3 } \right| – \left| {z_4 } \right|} \Big |}}
$$

I know that you can rewrite this as
$$
\left| {\frac{{z_1 + z_2 }}{{z_3 -( -z_4 )}}} \right| \le \frac{{\left| {z_1 } \right| + \left| {z_2 } \right|}}{{\Big | {\left| {z_3 } \right| – \left| {-z_4 } \right|} \Big |}}
$$
Further, I know that the triangular inequalities are:
abs=absolute value

1. $\operatorname{abs}(z_1+z_2)\leq \operatorname{abs}(z_1) +\operatorname{abs}(z_2)$
2. $\operatorname{abs}\big(\operatorname{abs}(z_1)-\operatorname{abs}(z_2)\big) \leq \operatorname{abs}(z_1-z_2)$

but I'm not sure how to put these ideas together

Answer 1

Since $|z_1+z_1|\leqslant|z_1|+|z_2|$ and $|z_3+z_4|=|z_3-(-z_4)|\geqslant\bigl||z_3|-|z_4|\bigr|$, you have$$\left|\frac{z_1+z_2}{z_3+z_4}\right|=\frac{|z_1+z_2|}{|z_3+z_4|}\leqslant\frac{|z_1|+|z_2|}{\bigl||z_3|-|z_4|\bigr|}.$$