How to Prove This Complex Inequality Involving Absolute Values?

[saint_n]

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

 

[Sirus]

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...

 

[saint_n]

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

 

[saint_n]

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?