Seemingly Innocent Inequality
k = 1 ∑ n x k 2 ≤ k = 1 ∑ n x k 2 1
Given that x 1 , x 2 , … , x n are positive reals whose sum is n , find the largest integer n such that the inequality above always holds true.
If you think all positive integers n make the inequality hold true, enter 0 as your answer.
The answer is 10.
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2 solutions
Very interesting problem, one of the best posted on Brilliant. Wow.
Yes, it is surprising.
How do you prove that for n < 11 , the above inequality is true for all positive Xi's?
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I had it in my solution but it got removed for some reason.
Use Vasc's RCF Inequality. First link on a google search will get you the paper.
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Is there an accessible solution to this problem somewhere on the Internet? I have ad box proofs which work for small n, but have not been able to find any reason why n=10 is the cutoff point. I am really curious about this remarkable result!
I don't get it. I assumed all x values to be equal to 1. Which would make the sum of x1 to xn equal to n. Left side of the equality would be equal to n. Right side would also be equal to n. Thus the inequality would hold for all positive values of n.
Tell me where I got it wrong.
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The inequality must always hold true; in other words, it must hold true no matter what x 1 , x 2 , … , x n you plug in (such that their sum is n ).
Very interesting,indeed
Not sure , but how about setting n-1 numbers to have value 1/(n-1) and last number to be equal to (n-1). In this case LHS is (n-1)^2 + 1/(n-1) while RHS is (n-1)^3+ 1/(n-1)^2. RHS is always greater than LHS in this case for all n > 2
I've got a pretty tedious proof using Lagrange Multipliers that 10 is indeed the answer, but it is very tedious. On a side note, u can further widen the inequality with
( x 1 , x 2 , . . . , x 1 1 ) = ( 0 . 6 2 0 7 3 6 2 9 3 4 4 , 0 . 6 2 0 7 3 6 2 9 3 4 4 0 , . . . , 0 . 6 2 0 7 3 6 2 9 3 4 4 , 4 . 7 9 2 6 3 7 0 6 5 5 8 )
Which gives the maximum difference between the RHS and LHS of the inequality to be 0.826079456408. This seems more of a calculus problem than it is a algebra one since there isn't nice numbers.
You need to put (n-1) numbers as n/(2(n-1)) and other as n/2 . The inequality then gives n<=6+2sqrt(6)<11
It's quite surprising that this inequality doesn't hold true for all values of n (or surprising that it does hold true at all). In fact, the inequality holds true for all positive integer values of n until n ≥ 1 1 , where a counterexample is ( x 1 , x 2 , … , x 1 1 ) = ( 3 2 , 3 2 , … , 3 2 , 3 1 3 ) This counterexample is mainly motivated by setting n − 1 variables equal to each other, and testing the inequality after that.