# Baby Rudin Proof of Theorem 1.33 (e) - Triangle Inequality

1. Jun 7, 2012

### josueortega

Hi everyone,

I have a question on Rudin's proof of Theorem 1.33 part e. Here he prove the following statement:

The absolute value of z+w is equal or smaller than the absolute value of z plus the absolute value of w -Yes, is the triangle inequality, where z and w are both complex numbers-
|z+w| $\leqslant$ |z| + |w|

In the proof, the key is that he points out that
$$2Re(z\overline{w}) \leqslant 2|z\overline{w}|$$

which obviously implies that
$$Re(z\overline{w}) \leqslant |z\overline{w}|$$

Why is that so? How does he knows this inequality is satified? If you can help me I would appreciate it a lot.

2. Jun 7, 2012

### micromass

It's from part (d)...

3. Jun 7, 2012

### josueortega

I think I got it now! We know that

$$|Re(x)| \leqslant |x|$$

$$[Re(x)Re(x)]^{1/2} \leqslant |x|$$

$$Re(x) \leqslant |x|$$ which is what we wanted to prove. Right?

4. Jun 11, 2012

### dylanbyte

I think that you are slightly overthinking this, the real part of a complex number is just that - a real number.

And the modulus of a real number is just its absolute value, and a real number is always less than its absolute value.