HELP Pls

Thank you for putting it in latex so its readable :) 

The basic idea when we see the word minimize is to think of inequalities. I suspect you got this problem from a book about inequalities. We also see squares, and that gives us the idea to use Cauchy Schwarz inequality. As a recap, it states \(\left| \sum_{i=1}^na_ib_i \right|^2 \le \left(\sum_{i=1}^{n}|a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right) \). See here: https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality?srsltid=AfmBOore-GDD9oAtWZDDSJ-URH43dfQd2atn26PRaLxAr01bGo2-r183 (aops wiki page). 

This indeed matches with minimizing squares. Now, the question becomes, what do we let \(a_i\) and \(b_i\) be? [Looking back on it, we shouldn't use \(a_i\)to avoid confusion, but oh well... our "\(a_i\)" here is refering to the Cauchy Schwarz inequality, and not to the problem.] Well, let's look at our coefficients. we have 1, 2, 3, ... , 20. Since we are squaring things in our formula, we probably want something in the form \(\sqrt{1}, \sqrt{2}, \sqrt{3}, ... , \sqrt{20} \). So if we have\(a_i = \sqrt{i} \implies \sum_{i=1} ^{20}{a_i} = \sqrt{1} + \sqrt{2} + ... + \sqrt{20} \implies \sum_{i=1} ^{20}{a_i^2} = 1 + 2 + ... + 20\). What about \(b_i\)? We have \(a_i\) in the form of some square roots, but when we want to minimize \(\left| \sum_{i=1}^na_ib_i \right|^2\), we have \(A_i\) terms and whole numbers (here, \(A_i\) stands for the real numbers given in the problem). So, we want a \(b_i\) such that when it is multiplied by \(a_i\), we get nice coefficients. It becomes obvious (at least to me), that \( \sum_{i=1} ^{20}{b_i} = \sqrt{1} A_1 + \sqrt{2}A_2 + \sqrt{3}A_3 ... + \sqrt{20} A_{20} \implies \sum_{i=1} ^{20}{b_i ^ 2} = 1 A_1^2 + 2A_2^2 + 3A_3^2 ... + 20 A_{20}^2. \) Then, \( \sum_{i=1} ^{20}{(a_i b_i)^2} = 1A_1 + 2A_2 + 3A_3 ... + 20 A_{20}\). 

Finally, putting it all together, we have \(1A_1 + 2A_2 + 3A_3 + ... 20A_{20} \le (1 + 2 + 3 + ... + 20) (1A_1^2 + 2A_2^2 + 3A_3 ^ 2 + ... + 20 A_{20}^2)\). From the problem, we have \(1A_1 + 2A_2 + 3A_3 + ... 20A_{20} = 1\), and we know the sum of \((1 + 2 + 3 + ... + 20) = 210\)(You can think of it as grouping: 1+20 = 21, 2+19 = 21, 3+18 = 21. We have 10 total pairs, so 21x10 = 210. This is the derivation of the formula: \(1 + 2 + ... + n = \frac{n(n-1)}{2}\)). ]

Plugging these values in, we have \(\frac{1}{210} \le (1A_1^2 + 2A_2^2 + 3A_3 ^ 2 + ... + 20 A_{20}^2)\). 

Now, to minimize this, we need equality to hold. The equality condition is as follows: 

We need some \(x\) such that \(x a_i = b_i\), or \(x \sqrt{i} = \sqrt{i} A_i\), so \(x = A_i\) for all \(i \in {(1, 2, 3, ..., 20)}\)[The "e" notation is reads "in", so for all i "in" (1, 2, ..., 20).] 

From the original problem, we have \(1A_1 + 2A_2 + 3A_3 + ... 20A_{20} = 1\), so plugging in \(x = A_i\) gives \(1 + 2x + 3x + ... 20 x = 1 \implies x(1+2+3 + ... + 20) = 1, \text{so } x = \frac{1}{120}\). So, \(a_{12} = 12 \cdot \frac{1}{120} = \frac{1}{10}\). 

Note: I may have made some mistakes on the way, so make sure to read the solution, understand everything, and see what mistakes I made... Because just getting the answer without learning doesn't help you