# Triangle inequality for complex numbers: sketch of proof

## Homework Statement

Show that if $z_1,z_2 \in \mathbb{C}$ then $|z_1+z_2| \leq |z_1| + |z_2|$

Above.

## The Attempt at a Solution

I tried by explicit calculation, with obvious notation for $a,b$ and $c$: my frist claim is not that the triangle inequality holds, just that I don't know to put a ? above the $\leq$ symbol

$\sqrt{a} \leq \sqrt{b} + \sqrt{c} \rightarrow 0 \leq \sqrt{b} + \sqrt{c} - \sqrt{a} \rightarrow 0 \leq \frac{b+c+2\sqrt{b}\sqrt{c} - a}{\sqrt{b}+\sqrt{c}+\sqrt{a}}$
Now if $z_1 = x_1 + i y_1$ and $z_2=x_2+iy_2$

and using again the conjugate of the roots expresion, the last equation is something like

$0\leq f(x_1^2,x_2^2,y_1^2,y_2^2)$

and so is true. can this be correct or may I write explicitly all the terms?

Thanks.

## Answers and Replies

I really don't think that's going anywhere. You should use complex conjugates to express the absolute values. $|z|^2=z \bar{z}$. Start from $|z_1+z_2|^2=(z_1+z_2) (\bar{z_1} + \bar{z_2})$