Proving an Inequality with Complex Numbers

[mjordan2nd]

Homework Statement

If z and w are complex numbers such that |z|<=1 and |w|<=1 then prove

$$\left| z+w \right| \leq \left| 1 + \overline{z} w \right|$$

The Attempt at a Solution

I have reduced this to essentially

x^2+y^2 <= 1+(xy)^2.

It seems to me if both x and y are less than or equal to 1, then the inequality must hold. I can't think of how to prove this formally, though. Any help on how to do this would be appreciated.

 

[Dick]

Ok, so if x^2=a and y^2=b then you want to prove a+b<=1+ab, where 0<=a<=1 and 0<=b<=1. Write it as a*(1-b)+1*b<=1. If b is in [0,1] then the left side is a number between 'a' and '1', right? There's a name for this kind of inequality, but I forget what it is. There may also be an easier way to prove this. But I forget that too.

 

[Dick]

Maybe it's a form of a Jensen's inequality, in kind of a vague way.

 

[mjordan2nd]

Dick,

Thanks for your post. I am sorry, but I do not see how/why the following is true:

If b is in [0,1] then the left side is a number between 'a' and '1', right?

 

[dirk_mec1]

Suppose $$|\bar{z}| < 1$$ then$$|w|^2 \le \frac{1-|z|^2}{1-|z|^2}$$$$|w|^2-|\overline{z} w|^2 \le 1-|z|^2$$$$|w|^2 + |z|^2 + 2\mbox{ Re}(\overline{z} w) \le 1 + |\overline{z} w|^2 + 2\mbox{ Re}(\overline{z} w)$$$$\left| z+w \right|^2 \leq \left| 1 + \overline{z} w \right|^2$$

 

[Dick]

Dick,

Thanks for your post. I am sorry, but I do not see how/why the following is true:

Can you show if t is in [0,1] then f(t)=a*t+b*(1-t) is between a and b? f(0)=b, f(1)=a and f(t) is a linear function.