inequality - Showing the "only if" direction of equality in a complex equality

How can I finish off the "only if" direction? I am just unable to prove the only if direction! Using the induction hypothesis and the triangle inequality is confusing me for some reason.

Show that
\begin{equation}
|z_1+z_2+\dots+z_n| = |z_1| + |z_2| + \dots + |z_n|
\end{equation}
if and only if $z_k/z_{\ell} \ge 0$ for any integers $k$ and $\ell$, $\ell \leq k, \ell \leq n,$ for which $z_{\ell} \ne 0.$

We show the "if" direction first. Suppose that $z_k/z_{\ell} \ge 0.$ Without loss of generality, suppose that $z_1$ is nonzero. Otherwise, we could reduce to $|z_2+\dots+z_n| = |z_2| + \dots + |z_n|$, where $z_2, \dots, z_n$ are all nonzero. Then we have:
\begin{align*}
|z_1+z_2+\dots+z_n| \ &= |z_1|\left|1+\dfrac{z_2}{z_1}+\dots+\dfrac{z_n}{z_1}\right| \\
&= |z_1|\left(1+\dfrac{z_2}{z_1}+\dots+\dfrac{z_n}{z_1}\right) \ \ \ \ \ \ \ \ \ \ \ \mathrm{Since} \ \dfrac{z_i}{z_1} \ge 0 \\

&=|z_1|\left(1+\left|\dfrac{z_2}{z_1}\right|+\dots+\left|\dfrac{z_n}{z_1}\right|\right) \\
&=|z_1|\left(1+\dfrac{|z_2|}{|z_1|}+\dots+\dfrac{|z_n|}{|z_1|}\right)=|z_1| + |z_2| + \dots + |z_n|
\end{align*}
To show the "only if" direction, we use induction. For $n=2$, we want to
find a condition for which $|z_1+z_2|=|z_1|+|z_2|.$ From the book and class discussions, we see that equality occurs if $z_1$ and $z_2$ are collinear. Provided the valid assumption of $z_2 \ne 0,$ we have that a necessary and sufficient condition, for which $|z_1+z_2|=|z_1|+|z_2|$, is $z_1/z_2 \ge 0.$

Thanks!

Answer

If $z_k/z_l$ is not $\ge 0$ for some integers $k$ and $l$, let $k=1$ and $l=2$ without loss of generality. You already know that $|z_1+z_2|<|z_1|+|z_2|$. Then
\begin{align*}

|z_1+z_2+\cdots+z_n|&\le |z_1+z_2|+|z_3|+\cdots+|z_n|\\
&<|z_1|+|z_2|+|z_3|+\cdots+|z_n|,
\end{align*}
which means $|z_1+z_2+\cdots+z_n|= |z_1|+|z_2|+\cdots+|z_n|$ does not hold. Q.E.D.

The triangle inequality, which I used in the first line, can be shown by induction easily.