Proving an inequality

1. Jan 21, 2010

mjordan2nd

1. The problem statement, all variables and given/known data

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

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

3. 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.

2. Jan 21, 2010

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.

3. Jan 21, 2010

Dick

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

Last edited: Jan 21, 2010
4. Jan 21, 2010

mjordan2nd

Dick,

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

5. Jan 22, 2010

dirk_mec1

Suppose $$|\bar{z}| < 1$$ then

$$|w|^2 \le \frac{1-|z|^2}{1-|z|^2}$$

$$|w|^2-|\overline{z} w|^2 \le 1-|z|^2$$

$$|w|^2 + |z|^2 + 2\mbox{ Re}(\overline{z} w) \le 1 + |\overline{z} w|^2 + 2\mbox{ Re}(\overline{z} w)$$

$$\left| z+w \right|^2 \leq \left| 1 + \overline{z} w \right|^2$$

Last edited: Jan 22, 2010
6. Jan 22, 2010

Dick

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.