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

[Jeff.Nevington]

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

∑_{n=0 to N-1} (|B_n|^x) ≥ Nα^x

where:
x > 1;
α = (1/N) * ∑_{n=0 to N-1} (|B_n|) (i.e., The mean of the absolute elements of B).
and ∑_{n=0 to N-1} (||B_n|-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.

 

[StoneTemplePython]

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)

 

[mfb]

This is a generalization of the inequality of arithmetic and geometric mean. Wikipedia has a discussion.

 

[Jeff.Nevington]

Thanks a lot! Jensen's Inequality, was what I was looking for. The Wiki page on mean inequality was also useful.