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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.

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  • $\begingroup$ Non-zero omplex numbers $\endgroup$ – Nosrati Jan 10 '17 at 9:57
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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.

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We have

(1) $Re(z+w)=Re(z)+Re(w) \le |z|+|w|$

and

(2) $||u|-|v|| \le |u+v|$.

Your turn !

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