Question

Help please! Medal and fan!

Answer 1

Assume that
\[
1a_1+2a_2+\cdots+na_n=1,
\]
where the a_j are real numbers.
As a function of $n$, what is the minimum value of
\[1a_1^2+2a_2^2+\cdots+na_n^2?\]

Answer 2

Sum of products => want to use Cauchy-Scharz ?

Answer 3

I know that I'm supposed to use Cauchy-Scharz, but I'm not sure how to apply it.

Answer 4

decompose \(ka_k^2\) in the right way... if you don't see right away, test them all!
Try: \(ka_k^2 = ka_k \times a_k\).

Answer 5

I see, it's tricky. Gotta think :D

Answer 6

Well before I had $$( 1a_1 + 2a_2 + 3a_3 + \cdots + na_n )^2 \le (1+2+3+ \cdots + n)(a_1^2 + a_2^2 + a_3^2 + \cdots + a_n^2 ),$$ and I simplified to $$ 1/(1+2+3+ \cdots + n) \le (a_1^2 + a_2^2 + a_3^2 + \cdots + a_n^2)., would this way also result in the correct answer?

Answer 7

Hmm I don't know. (you forgot "to the power 2" on the RHS).

The question itself is a hint. You want a lower bound, while usually CS gives you the upper bound. That means your expression is on the RHS of CS.
You must find \(c_k\) and \(d_k\) such that \((c_k d_k)^2\) can be written as \(ka_k^2 \times \star\).

Answer 8

So that \((\sum_k c_k d_k)^2 \le (\sum_k \star_k^2) (\sum_k ka_k^2)\).

Answer 9

Ummm...I'm not sure I know how to do that.

Answer 10

Find the things backwards.
CS: \((\sum_k |x_ky_k|)^2 \le (\sum_k x_k^2) (\sum y_k^2)\).

Here, the sum \(\sum_k y_k^2\) has terms \(ka_k^2\). That means \(y_k = \sqrt{k}a_k\).
Now find the "natural" complement \(x_k\) to this \(y_k\).

Answer 11

Sorry I probably seem kind of dumb but I still don't fully understand what I should do to solve the problem

Answer 12

In the resolution, there must be a guess. Turns out the natural guess works fine.

1. What I said above: if you want to find a lower bound for a sum, that means this sum is on the RHS of CS.
2. therefore the sum has terms "\(y_k^2\)". -> \(y_k^2\)
3. Find a "good" \(x_k\) and look at CS's inequality with this \(x_k,y_k\).

Answer 13

*2. "\(y_k^2\) -> \(y_k = \sqrt{k} \,a_k\).

Answer 14

If you multiply \(\sqrt{k}\,a_k\) by \(\sqrt{k}\), you obtain a term with a familiar face: \(ka_k\).
Try to write CS with this choice of \(x_k,y_k\): \((x_k,y_k)=(\sqrt{k},\sqrt{k}\,a_k)\).

Answer 15

Okay and once I do that what do I do?

Answer 16

Write CS:
\((\sum_k \sqrt k \sqrt k a_k)^2 \le (\sum_k (\sqrt k)^2)(\sum_k (\sqrt k a_k)^2)\)

Answer 17

This is almost the end.

Answer 18

Sorry it's just that I haven't been taught much about summations (except what they are called and what they look like).

Answer 19

ok i'll rewrite it

Answer 20

CS:
\[
(\sqrt 1\sqrt1a_1 +\sqrt 2\sqrt 2a_2+\dotsb+\sqrt n \sqrt na_n)^2 \\
\le \biggl((\sqrt 1)^2 + (\sqrt 2)^2 + \dotsb + (\sqrt n)^2\biggr) \biggl((\sqrt1a_1)^2+(\sqrt2a_2)^2+\dotsb+(\sqrt na_n)^2\biggr)
\]
Compute what's inside each the parnetheses.

Answer 21

Wait how do I compute what is in the parentheses if "n" could be any number?

Answer 22

You have the assumption for the LeftHandSide, and there is a "well-known" formula:
\(1+2+3+\dotsb+n = n(n+1)/2\).

Answer 23

Just a clarification for the LHS you are saying sqrt(n)*sqrt(n)*a_n not sqrt(n)*sqrt(n*a_n)? Right?

Answer 24

Yes. \(\sqrt{k} \times (\sqrt k \times a_k)\)

Answer 25

Do you just mean to simplify it:
LHS=$$(1a_1+2a_2+...+na_n)^2$$

Answer 26

That's it. The assumption now says that \(1a_1+\dots+na_n = 1\). -> \(1^2=1\) :)

Answer 27

Wait so is that the answer or is there more?

Answer 28

@reemii

Answer 29

1 isn't the correct answer though...

Answer 30

This is the LHS. Now:
CS: \(1^2 \le (1+2+\dotsb+n)\times (\text{your sum})\)
i.e.: \(\frac{1}{n(n+1)} \le \text{your sum} \).

Answer 31

because 1+2+...+n = n(n+1)/2.

Answer 32

typo..

\(\frac{2}{n(n+1)} \le \text{your sum}\)

Answer 33

Wait but how do I find the minimum of the sum. The inequality suggests I have to find the maximum of $$\frac{2}{n(n+1)}$$

Answer 34

I thought you wanted to find a minimal for \(1a_1^2+2_a_2^2+\dotsb +na_n^2\).

Answer 35

\(1a_1^2+2 a_2^2+\dotsb +na_n^2\)

Answer 36

oh yeah

Answer 37

Wait but then using the inequality how do I find the answer?

Answer 38

CS has given you this: \( 1 \le \text{sum} \times \text{yoursum}\). -> \(\text{yoursum}\ge 1/\text{sum} = 1/\frac{n(n+1)}{2}\).

Answer 39

= 2/(n(n+1)).

Answer 40

You're done

Answer 41

Oh. Thanks so much!