# |a_1 + a_2 + + a_n| <= sqrt(n)*sqrt((a_1)^2 + (a_2)^2 + + (a_n)^2)

## Homework Statement

For a natural number n and real numbers a1, a2,...,an, verify that:
|a1 + a2 + ... + an| <= sqrt(n)*sqrt(a12 + a22 + ... + an2)

## Homework Equations

I suspect that this can be done using properties of the inner product (i.e. the Cauchy-Schwarz inequality), or the triangle inequality, but I just can't seem to make it come out.

## The Attempt at a Solution

It obviously would be sufficient to prove that (a_1 + a_2 + ... + a_n)^2 <= n(a_1)^2 + n(a_2)^2 + ... + n(a_n)^2. But try as I might I can't figure a strategy to show this. The square of an n term sum of numbers is by no means pretty, and I don't have a good formula for it.

Just trying it out on paper, it seems like an induction proof. To get a feel for what you should do, play around with n = 1. It's a trivial equality. For n = 2, do some rearranging and then consider the AM-GM inequality.

Thank you for your reply. I got it figured out. It's essentially the RMS-AM inequality, which I realized by looking at the AM-GM inequality again and remembering that it is one part of a bigger statement. And happily, the RMS-AM part follows directly from the Cauchy-Schwarz inequality:

Cauchy tells us: |<x,y>| <= ||x|| ||y||
Let x=(a_1, a_2, ..., a_n), y=(1,1,...,1)

Then C-S says:
|a_1*1 + a_2*1 + ... + a_n*1| <= sqrt(a_1^2 + a_2^2 + ... + a_n^2)*sqrt(1+1+...+1) ==>
|a_1 + a_2 + ... + a_n | <= sqrt(n)*sqrt(a_1^2 + a_2^2 + ... + a_n^2)

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