I think part of my issue is that whenever I have seen the cauchy inequality, it is in the form of (u,v) for example, like to distinct vectors but I am confused on if we have one here or what?

Can you express what I have written on the RHS as an inner product? Think of the definition of the Euclidean inner product.

Hm, could I write the RHS as the inner product with itself?

Im still not exactly sure, is there any way you could elaborate on the solution? I appreciate the hints, but Im not sure if what I am saying is valid or not? Should I start with writing the RHS of the original equations as the inner product of the a's with themselves? and then the LHS and the inner product of u,v in the manner suggested by Timbuc?

@Timbuc okay so if I do that I would get $(a_1+..+a_n/\sqrt{n})$ ^2 $\le (a_1^{2}+…+a_n^{2})(1/$\sqrt{n}$+..+1/$\sqrt{n}$)? But how does this help?

@LearningMath You're not taking the norm of $(1/\sqrt{n},\ldots,1/\sqrt{n})$ correctly (and you need to square it, not that it matters in this particular case)...

Oh should it just be (1/n+…+1/n)?

Yes...which is equal to?

(1/ $\sqrt{n}$+…+1/ $\sqrt{n}$)^${2}$?

No, that's completely off. $(a^2 + b^2) \neq (a+b)^2$. Stop doing algebra and think: how many $1/n$'s are in that sum?

n of them, so it becomes 1?

@neuguy now I have $(a_1+..+a_n/ \sqrt{n})^{2} \le (a_1+..+a_n)^{2}$

Think about the hint I gave you at the start...and no, that's NOT what you have.

But is it valid to say $(a_1+..+a_n/ \sqrt{n})^{2} = (a_1+…+a_n)^{2}(1/\sqrt{n})^2$

Oops right I meant $ (a_1+…+a_n/ \sqrt{n})^2 \le (a_1^{2}+…+a_n^{2})$

At this point I'm not sure what you do not understand. Do you understand how to reach the conclusion of the proof from what we have worked out so far?

I think so but I am not sure, from what I gather we showed using the definition of the cauchy schwarz that we had originally was equivalent

But Im just not sure the best way to tie it together, what I have is starting with u=(a_1,…,a_n) and v=(1/sqrtn,..1/sqrtn), then using <u,v>^2 <= <u,u><v,v>

gives (a1+…+an/sqrt(n))^2 <= (a1^2+…+an^2)(1)

but now I'm just not sure if i should change the LHS to ((a1+…+an)^2)((1/sqrtn)^2)

which would give back the LHS, and I am not 100% if it is valid/circular

What is not valid or circular about it? We have used Cauchy-Schwarz on an entirely equivalent form of the LHS to draw this conclusion.

If you like, write it as a chain of equivalences and inequalities.