Proof of the generalised complex number triangle inequality without induction

Question

Does anyone have a proof of the generalised complex number triangle inequality that doesn't involve induction... is there one?

Thanks

Answer 1

We need to prove that $$|z_1|+|z_2|+...+|z_n|\geq|z_1+z_2+...+z_n|.$$ Indeed, let $z_k=a_k+b_ki,$ where $\{a_k,b_k\}\subset\mathbb R$.

Thus, by C-S $$\sum_{k=1}^n|z_k|=\sum_{k=1}^n\sqrt{a_k^2+b_k^2}=\sqrt{\sum_{k=1}^na_k^2+\sum_{k=1}^nb_k^2+2\sum_{1\leq i<j\leq n}\sqrt{(a_i^2+b_i^2)(a_j^2+b_j^2)}}\geq$$ $$\geq\sqrt{\sum_{k=1}^na_k^2+\sum_{k=1}^nb_k^2+2\sum_{1\leq i<j\leq n}\sqrt{(a_ia_j+b_ib_j)^2}}=$$ $$=\sqrt{\sum_{k=1}^na_k^2+\sum_{k=1}^nb_k^2+2\sum_{1\leq i<j\leq n}|a_ia_j+b_ib_j|}\geq$$ $$\geq\sqrt{\sum_{k=1}^na_k^2+\sum_{k=1}^nb_k^2+2\sum_{1\leq i<j\leq n}(a_ia_j+b_ib_j)}=$$ $$=\sqrt{\left(\sum_{k=1}^na_k\right)^2+\left(\sum_{k=1}^nb_k\right)^2}=|z_1+z_2+...+z_n|$$ and we are done!

Answer 2

Is this what you are looking for?

$$ |z_1 + \cdots + z_n|^2 = (z_1 + \cdots + z_n)(\bar z_1 + \cdots + \bar z_n) \\ = |z_1|^2 +\cdots + |z_n|^2 + \sum_{i < j}(z_i \bar z_j + \bar z_i z_j) \\ = |z_1|^2 +\cdots + |z_n|^2 + 2 \sum_{i < j} \operatorname{Re}(z_i \bar z_j ) \\ \le |z_1|^2 +\cdots + |z_n|^2 + 2 \sum_{i < j} |z_i z_j | = (|z_1| +\cdots + |z_n|)^2 \, . $$

Equality holds if $z_i \bar z_j$ is a non-negative real number for all $i, j$, that is if all non-zero $z_i$ are a positive multiple of each other.