Check my proof for equality in general triangle equality Consider the generalized triangle equality in $\mathbb C$:
$$|z_1+\dots+z_n|\le |z_1|+\dots+|z_n|\tag 1$$
The theorem in question is the following:

Equality occurs in (1) if and only if $z_k/z_l\ge 0$ for any integers
  $k,l$ where $1\le k,l\le n$ for which $z_l\ne 0$.

My proof is as follows:
The result is easy to verify for $n=2$. (This part is solved in the textbook). We proceed by induction and assume the result for $n-1$. If $z_n=0$ we are done anyway so we suppose $z_n\ne 0$. Let $z_{i_1},\dots,z_{i_k}$ be all the nonzero complex numbers amongst $z_1,\dots,z_{n-1}$. There must be at least one such nonzero number for otherwise the proof is trivial. Further let $z^*=z_{i_1}+\dots+z_{i_k}$. 
Now suppose equality prevails in (1). Then, $$|z_{i_1}|+\dots+|z_{i_k}|+|z_n|=|z^*+z_n|\le |z^*|+|z_n|$$ following which $|z^*|\ge |z_{i_1}|+\dots+|z_{i_k}|.\tag 2$ Also by the triangle inequality we have $|z^*|\le |z_{i_1}|+\dots+|z_{i_k}|.\tag 3$ 
From (2),(3) we see that $$|z^*|=|z_{i_1}|+\dots+|z_{i_k}|.$$ This shows that $z_k/z_l\ge 0$ for any integers $k,l$ where $1\le k,l\le n-1$ for which $z_l\ne 0$. Note that $z^*\ne 0$ for otherwise $z_{i_1}=\dots=z_{i_k}=0$.
Fix any $j$, $1\le j\le k$. By the induction hypothesis we now have $c_m=z_{i_m}/z_{i_j}>0$ where $1\le m\le k$. Since $|z^*+z_n|=|z^*|+|z_n|$ so, by the case $n=2$ of the result we have $z_n=tz^*$ for some $t>0$. Dividing both sides by $z_{i_j}$ we see that $$z_n/z_{i_j}=t(c_1+\dots+c_{j-1}+1+c_{j+1}+\dots+c_k)>0$$ following which $z_{i_j}/z_n>0$. This completes one direction of the proof.
Now suppose that for any $1\le p,q\le k$ we have $z_{i_p}/z_{i_q},z_{i_p}/z_n>0$. Then $\frac{z_{i_1}+\dots+z_{i_k}}{z_n}>0$ and so both $z^*/z_n,z+n/z^*>0$. Again, by the case $n=2$ this implies that $|z_n+z^*|=|z_n|+|z^*|$.
Hence,
\begin{align*}
|z_1+\dots+z_n|&=|z_{i_1}+\dots+z_{i_k}|+|z_n|\\
&=|z_{i_1}|+\dots+|z_{i_k}|+|z_n|\mbox{ (by the induction hypothesis)}\\
&=|z_1|+\dots+|z_n|
\end{align*}
which completes the proof.
Is this proof correct? Can it be improved?
 A: Here is a proof without induction:
In the following summations, all $z_i$ with $z_i =0$ are understood to be omitted.
Using  $|z|^2 = z z^*$ we have two proofs:


*

*First proof with polar complex variables:


$$
\vert{z_{1}+ \cdots + z_{n}}\vert^2 = (z_{1}+ \cdots + z_{n})(z_{1}^*+ \cdots + z_{n}^*) \\
= \sum_{i} |z_{i}|^2 +   \sum_{i > j} (z_i z_j^* +z_j z_i^* )\\
= \sum_{i} |z_{i}|^2 +  2 \sum_{i > j} \Re (z_i z_j^*) = \sum_{i} |z_{i}|^2 +  2 \sum_{i > j} |z_i||z_j| \cos(\phi_i - \phi_j)
$$
which must, if equality is to occur, be equal to
$$
(\vert{z_{1}}\vert + \cdots + \vert{z_{n}}\vert)^2 = \sum_{i} |z_{i}|^2 +  2 \sum_{i > j} |z_i||z_j|
$$
Since all terms in the last sum are positive and since $\cos(\phi_i - \phi_j) \le 1$, we must have  $\phi_i = \phi_j$ for all $(i,j)$. Hence all $z_j$ must have the same phase angle.  This is equivalent to  saying $z_{j} = c_{ji}z_{i}$ with real positive constants $c_{ji}$.


*Second proof with Re/Im complex variables:


In general, let $z_i = x_i + i y_i$
As in the first proof,
$$
\vert{z_{1}+ \cdots + z_{n}}\vert^2 
= \sum_{i} |z_{i}|^2 +  2 \sum_{i > j} \Re (z_i z_j^*) = \sum_{i} |z_{i}|^2 +  2 \sum_{i > j} (x_i x_j + y_i y_j)
$$
which must, if equality is to occur, be equal to
$$
(\vert{z_{1}}\vert + \cdots + \vert{z_{n}}\vert)^2 = \sum_{i} |z_{i}|^2 +  2 \sum_{i > j} |z_i||z_j|
$$
Now we look at each single term in the last sums. This can be immediately done by the Cauchy-Schwartz inequality. It states that 
$$
(x_i x_j + y_i y_j)^2 \le (x_i^2  + y_i^2 ) (x_j^2  + y_j^2 )  = |z_i|^2|z_j|^2
$$
Alternatively, you can interpret two real (!) vectors  $r_i = (x_i,y_i)$ and $r_j = (x_j,y_j)$ and refer to the simple scalar vector multiplication where it is well known that $r_i \cdot r_j \le |r_i||r_j|$ which gives the same inequality.
Since this holds for each term, the inequality is immediately established. However Cauchy-Schwartz (as well as scalar vector multiplication) has a more important issue to offer: it is known that  equality occurs if and only if the vector $(x_i,y_i)$ is a positive multiple of the  vector $(x_j,y_j)$. Again, this is equivalent to  saying $z_{j} = c_{ji}z_{i}$ with real positive constants $c_{ji}$.
