Show that for all n>=2 (x_1^2)/(x_1^2+x_2x_3)+(x_2^2)/(x_2^2+x_3x_4)+...+(x_(n-1}^2)/(x_(n-1)^2+x_nx_1)+(x_n^2)/(x_n^2+x_1x_2)<= n-1 where x_i are real positive numbers

Angel Kline 2022-10-19 Answered
Inequality x 1 2 x 1 2 + x 2 x 3 + x 2 2 x 2 2 + x 3 x 4 + + x n 1 2 x n 1 2 + x n x 1 + x n 2 x n 2 + x 1 x 2 n 1
Show that for all n 2
x 1 2 x 1 2 + x 2 x 3 + x 2 2 x 2 2 + x 3 x 4 + + x n 1 2 x n 1 2 + x n x 1 + x n 2 x n 2 + x 1 x 2 n 1
where x i are real positive numbers
I was going to use x 1 2 x 1 2 + x 2 x 3 = 1 1 + x 2 x 3 x 1 2 x 1 2 1 x 2 x 3 x 1 4 ( 1 x 2 + 1 x 3 )
x 1 2 x 1 2 + x 2 x 3 + x 2 2 x 2 2 + x 3 x 4 + . . . + x n 2 2 x n 2 2 + x n 1 x n + x n 1 2 x n 1 2 + x n x 1 + x n 2 x n 2 + x 1 x 2
= 1 1 + x 2 x 3 x 1 2 + 1 1 + x 3 x 4 x 2 2 + . . . + 1 1 + x n 1 x n x n 2 2 + 1 1 + x n x 1 x n 1 2 + 1 1 + x 1 x 2 x n 2
x 1 4 ( 1 x 2 + 1 x 3 ) + x 2 4 ( 1 x 3 + 1 x 4 ) + . . . + x n 2 4 ( 1 x n 1 + 1 x n ) + x n 1 4 ( 1 x n + 1 x 1 ) + x n 4 ( 1 x 1 + 1 x 2 )
= 1 4 ( ( x 1 x 2 + x 1 x 3 ) + ( x 2 x 3 + x 2 x 4 ) + ( x 3 x 4 + x 3 x 5 ) + . . . + ( x n 2 x n 1 + x n 2 x n ) + ( x n 1 x n + x n 1 x 1 ) + ( x n x 1 + x n x 2 ) )
= 1 4 ( ( x 1 + x 2 x 3 ) + ( x 2 + x 3 x 4 ) + ( x 3 + x 4 x 5 ) + . . . + ( x n 3 + x n 2 x n 1 ) + ( x n 1 + x n 2 x n ) + ( x 1 + x n x 2 ) + ( x n 1 + x n x 1 ) )
....I thought about using Cauchy's inequality, but that would only increase the problem
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Answers (1)

Kyle Delacruz
Answered 2022-10-20 Author has 21 answers
Let x 2 x 3 x 1 2 = a 1 a 2 ,... and similar, where a i > 0 and a n + 1 = a 1
Thus, we need to prove that:
i = 1 n 1 1 + a i a i + 1 n 1
or
i = 1 n ( 1 1 + a i a i + 1 1 ) 1
or
i = 1 n a i a i + a i + 1 1 ,
which is true because
i = 1 n a i a 1 + a i + 1 i = 1 n a i a 1 + a 2 + . . . + a n = 1
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