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)#?

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#.

#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.