if i have four consecutive numbers for example 5 6 7 8 i want to know why always sum of multiply every two consecutive numbers will always be greater than any combination else. for example ((8*7)+(6*5)) will always be greater than ((8*6)+(7*5)) and ((8*5)+(7*6)). i know it's an easy question but i want to be convinced
A product of two distinct integers is maximized when they are consecutive.
In your first example, case one, it happens two times in the same expression.
Your third example includes a product of a pair of consecutive integers, but
the sum is offset by a smaller product.
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If you know some algebra, you might see the following:
In increasing order, let the integers be n, n + 1, n + 2, n + 3
Case One)
n(n + 1) + (n + 2)(n + 3) =
n^2 + n + n^2 + 5n + 6 =
2n^2 + 6n + 6
. . . . . . . . . . . . . . . . . . . . .
Case Two)
n(n + 2) + (n + 1)(n + 3) =
n^2 + 2n + n^2 + 4n + 3 =
2n^2 + 6n + 3
. . . . . . .. . . . . . . ... . . . . . .
Case Three)
n(n + 3) + (n + 1)(n + 2) =
n^2 +3n + n^2 + 3n + 2 =
2n^2 + 6n + 2
_____________________
As you can see, the sum of the products for the
first case is always greater.
