Question:
Prove that: $$|z_1+z_2|^2 \le (1+c)|z_1|^2+(1+{1\over c})|z_2|^2$$
where $z_1,z_2$ are complex numbers and $c$ is a positive real parameter.
Solution:
We can write $$|z|^2=z\bar z$$
Same way $$(z_1+z_2)(\bar z_1+\bar z_2) \le (1+c)(z_1)(\bar z_1)+(1+{1\over c})(z_2)(\bar z_2)$$
which will yield $$z_2\bar z_1+z_1\bar z_2 \le cz_1\bar z_1+{1\over c}z_2\bar z_2$$
How to solve after this? Is this question correct? Am I doing correct?
Post script: Offtopic
This maybe irrelevant with the question. If you want to answer it you may or you may not.
As I'm quite new to the topic. Can anyone of you suggest a good book for a beginner as there is nothing left in the book I'm reading. It's just introduction with few problems. This is the last one.
How to pronounce $\bar z$? Some pronounce z bar and some strictly prohibit saying the same. They rather say complex conjugate of z.
Thanks for answering any of these questions or providing any hints.