Is this a theorem?
If x 1 , x 2 , x 3 , … , x n are any real number and n is any positiv integer, then
A.
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B.
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D.
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E. None of these
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2 solutions
In each choice I'll focus on just the simple case of n = 2. Of course, induction can be used for n > 2:
CASE (A): 2 * (x1^2 + x2^2) < x1^2 + 2*x1x2 + x2^2;
or x1^2 - 2*x1x2 + x2^2 < 0;
or (x1 - x2)^2 < 0 (contradiction!)
CASE (B): x1^2 + x2^2 >= x1^2 + 2*x1x2 + x2^2;
or 0 >= 2*x1x2, which only holds true when x1 and x2 are neither both positive or both negative, but not for any real x.
CASE (C): x1^2 + x2^2 >= 2 (x1^2 + 2 x1x2 + x2^2);
or 0 >= (x1^2 + 2 x1x2 + x2^2) + 2 x1x2;
or 0 >= (x1 + x2)^2 + 2 x1x2, which only holds true if (-2-sqrt(3)) x2 <= x1 <= (-2+sqrt(3))*x2, but not for any real x1.
CASE (D): Let the Sigma sums be represented as y, which gives:
y <= y^2,
or 0 <= y^2 - y;
or y <= 0 and y >= 1. Otherwise, we get a contradiction when 0 < y < 1.
CONCLUSION: None of these four Cases is valid for any real x
Your argument for case B is flawed since there's no reason why one of x1,x2 can't be negative and the other positive. For instance for n=2, x1=-1 and x2=2 the inequality holds and we get 5>=1. However it is still true that case B does not hold for all x. By C-S inequality there should be an n on the LHS of the inequality. If we choose for instance x1=x2=1/2 and n=2 we get 1/2>=1 which is false. Otherwise nice solution and great work in finding contradictions for all the cases!
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Thanks, Jonathan. Case (B) was pretty much the only one I struggled with articulating my proof. I did validate it only after trying a few test pairs for x1 and x2.
The answer follows simply from Cauchy Schwarz Inequality. Take the first series as x(i) and the second as 1(s). You can then easily cross check that none of the options match the condition.