Proving b_m ≥ ∑b_i^2 Under Given Conditions

[fmilano]

Hi, I need to show this:

b_m \geq\sum_{i=1}^n b_i^2

given these three conditions:

b_m \geq b_i, for i=1..n (in other words b_m = max(b_i)) and

0 \leq b_i \leq 1 for i=1..n and

\sum_{i=1}^n b_i=1

I've been working for hours in this without results...Any clue would be really appreciated

(this is not a homework exercise. I'm just trying to convince myself that the bayes decision error bound is a lower bound for the nearest neighbor rule error bound, and to convince myself of that I've arrived at the conclusion that I have to show the above).

Thanks,

Federico

 

[jgens]

Maybe this will help . . .

0 \leq b_{i}^2 \leq b_i \leq 1, this implies that \sum_{i = 1}^n b_{i}^2 \leq \sum_{i = 1}^n b_i = 1.