# Basic proof

1. Aug 4, 2009

### 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

2. Aug 4, 2009

### jgens

Maybe this will help . . .

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