Need help with a complex inequality??

[saint_n]

hey!!
i been trying to do this inequality for a 2 hrs now and cant 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 beacause you subtracting a ab on the RHS???