# Proof that involving inequalities and complex numbers

I'm trying to prove that, for complex numbers, $z_1, z_2, z_3, z_4$, where the modulus of $z_3$ is not equal to the modulus of $z_4$, that $$\frac{\text{Re\left(z_1 + z_2\right)}}{\lvert z_3 + z_4 \rvert} \leq \frac{\lvert z_1 \rvert + \lvert z_2 \rvert}{\left \lvert \lvert z_3 \rvert - \lvert z_4 \rvert \right \rvert},$$ where Re($z_1 + z_2$) denotes the real number of the sum of $z_1$ and $z_2$ and $\lvert z_1 \rvert$ denotes its modulus.

Here's my attempt at a proof of this.

First, we know that for some complex number, $z$, $\text{Re($z$)} \leq \left \lvert \text{Re($z$)} \right \rvert \leq \lvert z \rvert$. Thus, Re($z_1 + z_2$) $\leq \left \lvert z_1 + z_2 \right \rvert$. By the triangle inequality, $\left \lvert z_1 + z_2 \right \rvert \leq \lvert z_1 \rvert + \lvert z_2 \rvert$. Thus, by transitivity, Re($z_1 + z_2$) $\leq \lvert z_1 \rvert + \lvert z_2 \rvert$. It suffices now to demonstrate that $\left \lvert z_3 + z_4 \right \rvert \geq \left \lvert \lvert z_3 \rvert - \lvert z_4 \rvert \right \rvert$. This is, likewise, a consequence of the Triangle Inequality.

(If I'm not mistaken, the assumption that $\lvert z_3 \rvert \neq \lvert z_4 \rvert$ is only used to conclude that this second term has a non-zero denominator. If there's some other use here, please let me know.)

How does this look as a sketch of a proof? Many of this properties are already proved in the text, so I didn't see it as necessary to re-derive them. I suppose my central question here is whether, assuming the properties that I have mentioned, whether this fact follows.

Thanks.

• Looks like a proof to me. Jul 4 '18 at 16:44
• To me also.......
– Aqua
Jul 4 '18 at 16:49
• Actually, $\lvert z_3 + z_4 \rvert \geq \bigl\lvert \lvert z_3 \rvert - \lvert z_4 \rvert \bigr\rvert$ is standard, and is known as the triangle inequality, 2nd form. Jul 4 '18 at 16:55

$$\text{Re}\left(z_1 + z_2\right)\le\lvert z_1 + z_2 \rvert\le |z_1| +| z_2|\tag{1}$$
$$\left\lvert \lvert z_3 \rvert - \lvert z_4 \rvert \right\rvert\le\left \lvert z_3 + z_4 \right \rvert\implies \frac{1}{|z_3+z_4|}\le\frac{1}{||z_3|+|z_4||},\quad|z_3|\neq|z_4|\tag{2}$$ Now combine $(1)$ and $(2)$: $$\frac{\text{Re}\left(z_1 + z_2\right)}{\lvert z_3 + z_4 \rvert} \leq \frac{\lvert z_1 \rvert + \lvert z_2 \rvert}{\lvert z_3 + z_4 \rvert} \leq \frac{\lvert z_1 \rvert + \lvert z_2 \rvert}{\left \lvert \lvert z_3 \rvert - \lvert z_4 \rvert \right \rvert}$$