# prove the following inequality of complex number

Let $z,w,u,v$ be complex numbers.

Show that $\frac{Re(z+w)}{|u+v|} \leq\ \frac{|z| + |w|}{||u| - |v||}$

Ignoring the denomintor, the inequality of the numerator is the consequence of the triangle inequality of complex number, but how to I proceed on from here ?

Any help or insights is deeply appreciated.

• Non-zero omplex numbers – Nosrati Jan 10 '17 at 9:57

The inequality for the denominators also follows from the triangle inequality. Let $u,v\in \mathbf C$. We have $$\def\abs#1{\left|#1\right|} \abs{u} \le \abs{u-v}+ \abs{v} \iff \abs u - \abs v \le \abs{u-v}$$ and $$\abs{v} \le \abs{v-u} + \abs{u} \iff \abs v - \abs u \le \abs{v-u}$$ As $\abs{u-v}= \abs{v-u}$ both equations together imply $$\abs{\abs u - \abs v} \le \abs{u-v }$$ the so called reverse triangle inequality.
(1) $Re(z+w)=Re(z)+Re(w) \le |z|+|w|$
(2) $||u|-|v|| \le |u+v|$.