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)?

2 Answers
Sep 10, 2016

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#.

Sep 10, 2016

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.