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Feinsn

Feinsn

Answered question

2022-06-16

Let x 1 , . . . , x 25 > 0 be such that i = 1 25 x i = 4350 and i = 1 25 x i 2 = 757770.25.
From the first equality alone, we know that at least one of the x i 's must be less than or equal to 4350 25 = 174. From the second equality alone, we know that at least one of the x i 's must be less than or equal to 757770.25 25 = 174.1, which is less useful than the first bound. My question is whether we can get a better bound, i.e. to find the least upper bound of min { x 1 , . . . , x 25 } , when we use both equalities together. I appreciate any comments or hints.

Answer & Explanation

mallol3i

mallol3i

Beginner2022-06-17Added 20 answers

Here's some nontrivial improvement at least. Suppose that m = min { x 1 , , x 25 } 174, and write x j = m + t j . Then x j = 4350 means t j = 4350 25 m, while 757770.25 = x j 2 = ( m 2 + 2 m t j + t j 2 ) means
t j 2 = 757770.25 25 m 2 2 m t j = 757770.25 25 m 2 2 m ( 4350 25 m ) = 757770.25 8700 m + 25 m 2 .
Finally, the fact that the t j are all nonnegative means that ( t j ) 2 t j 2 ; in other words, we must have
( 4350 25 m ) 2 ( 757770.25 8700 m + 25 m 2 ) 0.
But now the quadratic equation implies that m 20880 59 6 120 , which is a little less than 172.8.
Edited to add: as RobPratt pointed out, this bound is actually optimal! If x 1 = = x 24 = 20880 59 6 120 and x 25 = 4350 24 x 1 = 870 + 59 6 5 , then x j = 4350 and x j 2 = 757770.25 exactly.

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