# Triangle inequality

1. Apr 12, 2006

### stunner5000pt

Edit: Nevermind i got it, thanks anyway

for complex numbers z1 and z2

prove that $$|z_{2}| - |z_{1}| \leq |z_{2} - z_{1}|$$

the left hand side becomes
$$\sqrt{x_{2}^2 + y_{2}^2} - \sqrt{x_{1}^2 + y_{1}^2}$$

the right hand side becomes
$$\sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}$$

now i tried squaring both sides and i get
left hand side
$$x_{2}^2 + y_{2}^2 + x_{1}^2 + y_{1}^2 - 2 \sqrt{(x_{2}x_{1})^2 + (x_{2}y_{1})^2 + (x_{1}y_{2})^2 + (y_{1}y_{2})^2}$$

right hand side
$$(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2$$
i put the two of them not equal to eah other and reduce and i ended up with
$$x_{2}^2 y_{1}^2 + x_{1}^2 y_{2}^2 \neq 2x_{1} x_{2} y_{1} y_{2}$$
im stuck now...

is there a simpler... less tedious way of doing this... by the way?

Last edited: Apr 12, 2006
2. Apr 12, 2006

### Galileo

The less tedious way is to use the triangle inequality: $|z-w|\leq |z|+|w|$