How to Prove This Complex Inequality Involving Absolute Values?

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Homework Help Overview

The discussion revolves around proving a complex inequality involving absolute values, specifically the expression |\frac{1}{2}(a+b)|^p \leq \frac{1}{2}(|a|^p+|b|^p, where a and b are complex numbers.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore various approaches to proving the inequality, with one suggesting a manipulation involving the term (\sqrt[p]{\frac{1}{2}}|(a+b)|)^{p} and questioning the validity of subtracting terms like pab. Others raise concerns about specific cases, such as when p = 1, and whether the manipulations hold true.

Discussion Status

The discussion is ongoing, with participants offering different lines of reasoning and questioning the assumptions behind the proposed manipulations. There is no clear consensus yet, as some participants express skepticism about the validity of certain steps.

Contextual Notes

Participants are considering the implications of different values of p and how they affect the inequality, particularly in the case of p = 1. There is also an acknowledgment of the complexity involved in handling absolute values in the context of complex numbers.

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Need help with a complex inequality??

hey!
i been trying to do this inequality for a 2 hrs now and can't seem to prove it
|\frac{1}{2}(a+b)|^p \leq \frac{1}{2}(|a|^p+|b|^p) where a,b are complex numbers
Can anyone suggest a way??
thanks
 
Last edited:
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Try this:

|\frac{1}{2}(a+b)|^{p}\leq(\sqrt[p]{\frac{1}{2}}|(a+b)|)^{p}-pab

Yes, I know, the last term is only pab if p=2, but you will always be subtracting somthing at the end, no matter the value of p. I think this kinda works...
 
so you saying that |\frac{1}{2}(a+b)|^{p}\leq(\sqrt[p]{\frac{1}{2}}|(a+b)|)^{p}-pab \leq \frac{1}{2}(|a|^p+|b|^p)
 
for p = 1
|\frac{1}{2}(a+b)|\leq(\frac{1}{2}|(a+b)|)-ab
isnt this false because you subtracting a ab on the RHS?
 
Last edited:

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