I How Can Jensen's Inequality Be Used to Prove a Vector Magnitude Relationship?

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I have a vector B of length N, I would like to prove that:

n=0 to N-1 (|Bn|x) ≥ Nαx

where:
x > 1;
α = (1/N) * ∑n=0 to N-1 (|Bn|) (i.e., The mean of the absolute elements of B).
and ∑n=0 to N-1 (||Bn|-α|) ≠ 0; (i.e., The absolute elements of B are not all identical).

I believe the above to be true, but I am struggling to find the most elegant way to state it.
 
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notationally this isn't great, but it seems that you have a function ##f##

where ##f(y) = y^\alpha##

for real non-negative ##y## and ##\alpha \gt 1##. Differentiate twice and show second derivative to be positive for any ##y\gt 0## (this is an easy way to confirm strict convexity).

If you divide both sides by ##N## then your inequality can be stated as

##E\Big[f\big(Y\big)\Big] \geq f\Big(E\big[Y\big]\Big)##

by Jensen's Inequality
with equality iff ##y_1 = y_2 = ... = y_n##

- - - -
edit: it's probably better to view this as a specific case of a Power Mean Inequality (which you'd in turn prove with help of convexity)
 
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This is a generalization of the inequality of arithmetic and geometric mean. Wikipedia has a discussion.
 
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Thanks a lot! Jensen's Inequality, was what I was looking for. The Wiki page on mean inequality was also useful.
 
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