An Elementary Inequality ProblemGiven n positive real numbers x 1 , x 2 , ....

Ishaan Booker

Ishaan Booker

Answered

2022-07-20

An Elementary Inequality Problem
Given n positive real numbers x 1 , x 2 , . . . , x n + 1 , suppose:
1 1 + x 1 + 1 1 + x 2 + . . . + 1 1 + x n + 1 = 1
Prove:
x 1 x 2 . . . x n + 1 n n + 1
I have tried the substitution
t i = 1 1 + x i
The problem thus becomes:
Given
t 1 + t 2 + . . . + t n + 1 = 1 , t i > 0
Prove:
( 1 t 1 1 ) ( 1 t 2 1 ) . . . ( 1 t n + 1 1 ) n n + 1
Which is equavalent to the following:
( t 2 + t 3 + . . . + t n + 1 t 1 ) ( t 1 + t 3 + . . . + t n + 1 t 2 ) . . . ( t 1 + t 2 + . . . + t n t n + 1 ) n n + 1
From now on, I think basic equality (which says arithmetic mean is greater than geometric mean) can be applied but I cannot quite get there. Any hints? Thanks in advance.

Answer & Explanation

frisiao

frisiao

Expert

2022-07-21Added 13 answers

By AM-GM
i = 1 n + 1 x i x i + 1 = i = 1 n + 1 k i 1 x k + 1 i = 1 n + 1 n k i ( 1 + x k ) n = n n + 1 i = 1 n + 1 ( 1 + x i )
and we are done!
enmobladatn

enmobladatn

Expert

2022-07-22Added 6 answers

From last, let t 1 t 2 . . . t n + 1 n = A with A M G M
t 2 + t 3 + . . . + t n + 1 n t 2 t 3 . . . t n + 1 n = A t 1 n
so for the first
t 2 + t 3 + . . . + t n + 1 n A t 1 n
and for the rest take like this.

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