When does $|z_1+z_2|^2 \leq |z_1|^2+|z_2|^2$ hold? When $z_1$ and $z_2$ are complex numbers, show with proof when does
$|z_1+z_2|^2 \leq |z_1|^2+|z_2|^2$ hold?
I guess that I can use $|z|^2=z\bar{z}$ with a triangle inequality
 A: If $z = x + yi$, where $x$ and $y$ are real, then I assume the absolute value signs means the standard modulus, such as defined in the Complex numbers section of Wikipedia's "Absolute values" article, i.e.,
$$|z|^2 = x^2 + y^2 \tag{1}\label{eq1A}$$
However, in that case, your statement is not always true. For example, have
$$z_1 = 2 + i \tag{2}\label{eq2A}$$
$$z_2 = 1 + 2i \tag{3}\label{eq3A}$$
Then
$$\begin{equation}\begin{aligned}
|z_1 + z_2|^2 & = |(2 + i) + (1 + 2i)|^2 \\
& = |3 + 3i|^2 \\
& = 9 + 9 \\
& = 18
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
but
$$\begin{equation}\begin{aligned}
|z_1|^2 + |z_2|^2 & = |2 + i|^2 + |1 + 2i|^2 \\
& = (4 + 1) + (1 + 4) \\
& = 10
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
As you can see, your statement of
$$|z_1+z_2|^2 \leq |z_1|^2+|z_2|^2 \tag{6}\label{eq6A}$$
is not true in this situation.
If you are using a different meaning for the absolute value signs, please specify what it is.
A: That is incorrect,
$$
\begin{aligned}
\arg{-\frac{z_{1}}{z_{2}}}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]&\iff \left|z_{1}+z_{2}\right|^{2}\leq|z_{1}|^{2}+|z_{2}|^{2}
\end{aligned}
$$
Maybe You mean 
$$
\left|\frac{z_{1}+z_{2}}{2}\right|^{2}\leq\frac{|z_{1}|^{2}+|z_{2}|^{2}}{2}
$$
A: This is false even for $z_1=z_2\ne 0$
$4|z_1|^2\le 2|z_1|^2$ doesn't hold.
A: $$|z_1+z_2|^2=(z_1+z_2)(\bar z_1+ \bar z_2)=z_1\bar z_1+\bar z_1 z_2+z_1 \bar z_2+ z_2 \bar z_2)= |z_1|^2+|z_2|^2+2 \Re (z_1 \bar z_2)$$
$$\implies |z_1+z_2|^2-|z_1|^2+|z_2|^2=2 \Re (z_1 \bar z_2)$$
Finally, iff $\Re (z_1 \bar z_2) \le 0$, then only we can have
$$|z_1+z_2|^2\le |z_1|^2+|z_2|^2, ~~if~~\Re (z_1 \bar z_2) \le 0$ $$
For example: $z_1=3-4i, z_2=3-5i$, then LHS is 185 and RHS is 99, becaues $\Re(z_1 \bar z_2)= +43.$
