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datomerki8a5yj

datomerki8a5yj

Answered question

2022-05-16

Let:
f ( x 1 , , x n ) = i x i ( 1 x i ) i < j | x i x j |
Suppose all x i ( 0 , 1 ) are fixed and i x i < n 2 . Show that there is some i and a sufficiently small ϵ so that x i x i + ϵ doesn't decrease the value of f.
That is to say, at least one of the partial derivatives of f is non-negative.
After taking the derivative in each x i one gets a system of inequalities. I was able to prove the statement for n = 2 , 3 this way through basically brute force. This doesn't generalize well though.

Answer & Explanation

necrologo9yh43

necrologo9yh43

Beginner2022-05-17Added 23 answers

It is actually not that difficult. Assuming that all xi are distinct and taking the logarithmic derivatives, we just need to show that at least one of the expressions
D i = 1 x i 1 1 x i + j : j i 1 x i x j
is positive. Now consider
i x i ( 1 x i ) D i = i ( 1 2 x i ) + i , j : i j x i x i 2 x i x j = σ + Σ .
We have σ = n 2 i x i > 0 by the assumption. Now we can use the antisymmetry of the denominator x i x j to write
2 Σ = i , j : i j ( x i x i 2 ) ( x j x j 2 ) x i x j = i , j : i j ( 1 x i x j ) = n ( n 1 ) 2 ( n 1 ) i x i > 0
and we are done.
Is it just an exercise from some book or you really needed it for something?
London Ware

London Ware

Beginner2022-05-18Added 3 answers

That was clean. I've was stuck on it for a while, didn't think it had an easy solution

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