Thanks for helping!
Let a1,a2,...,a20 be real numbers such that a1+2a2+⋯+20a20=1,
and a21+2a22+⋯+20a220 is minimized. Find a12
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 |∑ni=1aibi|2≤(∑ni=1|a2i|)(∑ni=1|b2i|). 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 ai and bi be? [Looking back on it, we shouldn't use aito avoid confusion, but oh well... our "ai" 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 √1,√2,√3,...,√20. So if we haveai=√i⟹∑20i=1ai=√1+√2+...+√20⟹∑20i=1a2i=1+2+...+20. What about bi? We have ai in the form of some square roots, but when we want to minimize |∑ni=1aibi|2, we have Ai terms and whole numbers (here, Ai stands for the real numbers given in the problem). So, we want a bi such that when it is multiplied by ai, we get nice coefficients. It becomes obvious (at least to me), that ∑20i=1bi=√1A1+√2A2+√3A3...+√20A20⟹∑20i=1b2i=1A21+2A22+3A23...+20A220. Then, ∑20i=1(aibi)2=1A1+2A2+3A3...+20A20.
Finally, putting it all together, we have 1A1+2A2+3A3+...20A20≤(1+2+3+...+20)(1A21+2A22+3A23+...+20A220). From the problem, we have 1A1+2A2+3A3+...20A20=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=n(n−1)2). ]
Plugging these values in, we have 1210≤(1A21+2A22+3A23+...+20A220).
Now, to minimize this, we need equality to hold. The equality condition is as follows:
We need some x such that xai=bi, or x√i=√iAi, so x=Ai for all i∈(1,2,3,...,20)[The "e" notation is reads "in", so for all i "in" (1, 2, ..., 20).]
From the original problem, we have 1A1+2A2+3A3+...20A20=1, so plugging in x=Ai gives 1+2x+3x+...20x=1⟹x(1+2+3+...+20)=1,so x=1120. So, a12=12⋅1120=110.
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