Question

Given `a in RR^+, a ne 1` and `n in NN, n > 1` Prove that `n^2 < (a^n + a^(-n)-2)/(a+a^(-1)-2)`?

Answer 1

Let a > 1. Then

`(a^n+a^(-n)-2)/(a+a^(-1)-2`

`(a^(-n)/a^(-1))((a^(2n)-2a^n+1)/(a^2-2a+1))`

`=1/a^(n-1)((a^n-1)/(a-1))^2`

`=1/a^(n-1)(1+a+a^2+a^3+...+a^(n-1))^2`

`>(1+1+1+....+1)^2/a^(n-1)`

, using `a^r>1, r=1. 2. 3, ...`

Likewise, when a < 1,

`(a^n+a^(-n)-2)/(a+a^(-1)-2`

`(a^n/a)((a^(-2n)-2a^( -n )+1)/(a^(-2)-2a^(-1)+1))`

`=a^(n-1)((1-a^(-(n-1)))/(1-a^(-1)))^2`

`=a^(n-1)(1+1/a+1/a^2+1/a^3+...+1/a^(n-1))^2`

`>a^(n-1)(1+1+1+....+1)^2`

`>n^2a^(n-1`

, using `a^(-r)>1, r=1. 2. 3, ...`

So, the given expression is

`> n^2/a^(n-1)`, for `a > 1` and

`> n^2a^(n-1)`, for a < 1#.

Answer 2

`a^n + a^(-n)-2=(a^(n/2)-a^(-n/2))^2`

then

`(a^n + a^(-n)-2)/(a+a^(-1)-2)=((b^n-b^(-n))/(b-b^(-1)))^2`

with `b = sqrt(a)` so

`n < (b^n-b^(-n))/(b-b^(-1))` calling now `b = e^lambda` we have

`n < ((e^lambda)^n-(e^lambda)^(-n))/(e^lambda-(e^lambda)^(-1)) = (e^(lambda n)-e^(-lambda n))/(e^lambda-e^(-lambda))=sinh(lambda n)/sinh(lambda)`

This is a continuous even function having a minimum at `lambda=0`

also we have

`lim_{lambda->0}(sinh(lambda n)/sinh(lambda)) = n`

so the proof follows.