If $z^2 = z_1^2+z_2^2+z_3^2$ then $|\Re(z)| \le|\Re(z_1)|+|\Re(z_2)|+|\Re(z_3)|$ Suppose $z,z_1,z_2,z_3 \in \mathbb{C}$ such that $z^2 = z_1^2+z_2^2+z_3^2$.
Prove that $$|\Re(z)| \le|\Re(z_1)|+|\Re(z_2)|+|\Re(z_3)|$$
where $\Re(z)$ is the real part of $z$.
I think we can use the triangle inequality but I can't complete it.
 A: First note that for all complex numbers $z$
$$ \tag{*}
\lvert\operatorname{Re}(z)\rvert^2 = \frac 14 (z + \overline z)^2 = 
 \frac 14 (z^2 + \overline z^2 + 2\lvert z\rvert^2) = 
\frac 12 \bigl(\operatorname{Re}(z^2)  + \lvert z^2 \rvert\bigr) \quad .
$$
Now we can prove the inequality for an arbitrary number of
complex numbers. If
$$
 z^2 = z_1^2 + z_2^2 + \ldots + z_n^2 
$$
then
$$
\begin{aligned}
 \lvert\operatorname{Re}(z)\rvert^2 &= \frac 12 \bigl(
  \sum_{k=1}^n \operatorname{Re}(z_k^2) +  \bigl\lvert \sum_{k=1}^n z_k^2  \bigr\rvert
 \bigr) \\
&\le \frac 12 \bigl(
  \sum_{k=1}^n \operatorname{Re}(z_k^2) +  \sum_{k=1}^n \lvert z_k^2 \rvert
 \bigr) \quad &\text{(triangle inequality)} \\
 &= \sum_{k=1}^n \lvert\operatorname{Re}(z_k)\rvert^2
\quad &\text{(using $(*)$ for all $z_k$)}
 \\
 &\le \bigl(
\sum_{k=1}^n \lvert\operatorname{Re}(z_k) \rvert 
 \bigr)^2
\end{aligned}
$$
and therefore
$$ \tag{**}
\lvert\operatorname{Re}(z)\rvert 
\le \sum_{k=1}^n  \lvert \operatorname{Re}(z_k) \rvert \, .
$$
When does equality hold?


*

*Equality in the first inequality (triangle inequality)
holds if and only if all $z_k^2$ are real, non-negative multiples of some $z_j^2$,
which means that all $z_k$ are real multiples of $z_j$.

*Equality in the second inequality holds if and only if
$\operatorname{Re}(z_k) \ne 0$ for at most one $k$.


Together it follows that equality holds in $(**)$ if and only if


*

*all $z_k$ are purely imaginary, or

*$z_k \ne 0$ for at most one $k$.


In the first case, $z$ is purely imaginary as well. 
In the second case, $z = \pm z_k$.
