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## Homework Statement

Reading Feynman The Principle of Least Action out of The Feynman Lectures on Physics, Vol 2. Link to text http://www.feynmanlectures.caltech.edu/II_19.html

So I'm having a problem proving that, section 19-2 5th paragraf, that

"Now the mean square of something that deviates around an average, as you know, is always greater than the square of the mean; ". The purpose of this was to illustrate the principle of least action. Then proving that the path that is taken by a particle is the one with Actions = 0, and the way to get to that path. Below is a comment that is underlined(sp?) made while discussing how to do this.

## Homework Equations

So definitions:

Numbers: [tex] x_1+...x_n [/tex]

Mean: [tex] (x_1+...x_n)/n [/tex]

Mean square: [tex] ((x_1+...x_n)/n)^2 [/tex]

Square mean: [tex] (x^2_1+...x^2_n)/n [/tex]

Kinetic Energy: KE

Potential Energy: PE

Action [tex] = S = \int_{t_1}^{t_2} (KE-PE) dt [/tex]

## The Attempt at a Solution

So i can prove for myself that the following is true:

[tex] ((x_1+...x_n)/n)^2 \geq (x^2_1+...x^2_n)/n \quad \forall x_i \geq 0 [/tex]

The problem here s that it is not always that [tex] x_i \geq 0 [/tex]. Since we have [tex] v \in \mathbb{R} [/tex], and therefore can be less then 0. So when we later go to find the path that is right we can't minimize with respect to the action.

What am I missing here?

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