Inspired by Aareyan Manzoor, again
S = k = 1 ∑ 2 0 1 6 x k x k + 1
Find the maximum of S if k = 1 ∑ 2 0 1 6 x k 2 = 2 0 1 6 and x 2 0 1 7 = x 1 .
The answer is 2016.
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2 solutions
same here(+1). this turns out to be easier than the original problem.
@Otto Bretscher , i dont remember "inspiring" any of you previous problems.
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It becomes harder if you omit the "link" x 2 0 1 6 x 1 . Try it!
I have now written a few problems on this topic @Aareyan Manzoor Enjoy!
A slight variant for those who dislike Cauchy-Schwarz (all sums run from 1 to 2016): 0 ≤ ∑ ( x k − x k + 1 ) 2 = ∑ x k 2 − 2 ∑ x k x k + 1 + ∑ x k + 1 2 = 2 ∑ x k 2 − 2 ∑ x k x k + 1 so ∑ x k x k + 1 ≤ ∑ x k 2 = 2 0 1 6
You still have to show that the this value is achievable.
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It seems self-evident from the inequality 0 ≤ ∑ ( x k − x k + 1 ) 2
;)
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It's not that self-evident.
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@Pi Han Goh – What about this one , Comrade Pi Han Goh?
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@Otto Bretscher – Must I use quadratic forms to get the answer too?
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@Pi Han Goh – Do it any way you can, Comrade... "Das Wesen der Mathematik liegt in ihrer Freiheit" (Cantor)
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@Otto Bretscher – What I mean to ask is "What is the solution you had in mind?" Classical inequalities? Highly doubt it. Lagrange Multipliers? don't think so.
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@Pi Han Goh – Classical inequalities are too weak for a problem like this , I believe (maybe somebody will prove me wrong). As you know, quadratic forms and Lagrange multipliers are equivalent in this case; either way you have to think about a matrix. This may require a bit of research: Find out what is known about the eigenvalues of a matrix "of this form". You can do it, Comrade!
Applying Cauchy Shwarz inequality, S = k = 1 ∑ 2 0 1 6 x k x k + 1 ≤ k = 1 ∑ 2 0 1 6 ( x k ) 2 k = 1 ∑ ( x k + 1 2 0 1 6 ) 2 = 2 0 1 6 2 = 2 0 1 6 ( S i n c e k = 1 ∑ 2 0 1 6 ( x k ) 2 = k = 1 ∑ 2 0 1 6 ( x k + 1 ) 2 = 2 0 1 6 ) Equality occurs when x 2 x 1 = x 3 x 2 = . . . . . = x 2 0 1 7 x 2 0 1 6 Or x 1 = x 2 = . . . . = x 2 0 1 6 = 1