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)#?

#(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, ...#

#(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, ...#

#> 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#

#(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#

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

To prove \(n^2 < \frac{a^n + a^{-n} - 2}{a + a^{-1} - 2}\) given \(a \in \mathbb{R}^+\), \(a \neq 1\), and \(n \in \mathbb{N}\), \(n > 1\): 1. First, let's define \(f(a) = a^n + a^{-n}\). We know that \(f(a)\) is strictly increasing for \(a > 0\) and \(a \neq 1\). 2. Now, let's consider the function \(g(a) = a + a^{-1}\). This function is also strictly increasing for \(a > 0\) and \(a \neq 1\). 3. Given that both \(f(a)\) and \(g(a)\) are strictly increasing, we can apply Jensen's inequality, which states that for a convex function \(f(x)\), if \(x_1, x_2, ..., x_n\) are non-negative real numbers and \(w_1, w_2, ..., w_n\) are positive real numbers such that \(w_1 + w_2 + ... + w_n = 1\), then: \[f(w_1x_1 + w_2x_2 + ... + w_nx_n) \leq w_1f(x_1) + w_2f(x_2) + ... + w_nf(x_n)\] 4. Let \(w_1 = \frac{1}{2}\) and \(w_2 = \frac{1}{2}\). We have: \[w_1 + w_2 = \frac{1}{2} + \frac{1}{2} = 1\] 5. Applying Jensen's inequality to \(f(a)\) and \(g(a)\), we get: \[f\left(\frac{a + a^{-1}}{2}\right) \leq \frac{1}{2}f(a) + \frac{1}{2}f(a^{-1})\] \[g\left(\frac{a + a^{-1}}{2}\right) \leq \frac{1}{2}g(a) + \frac{1}{2}g(a^{-1})\] 6. Substituting \(f(a)\) and \(g(a)\) into the inequality we want to prove: \[\frac{a^n + a^{-n}}{a + a^{-1}} \leq \frac{1}{2}(a^n + a^{-n}) + \frac{1}{2}(a + a^{-1}) - 1\] 7. Simplify the right-hand side: \[\frac{1}{2}a^n + \frac{1}{2}a^{-n} + \frac{1}{2}a + \frac{1}{2}a^{-1} - 1\] 8. Now, we need to prove that \(n^2 < \frac{1}{2}a^n + \frac{1}{2}a^{-n} + \frac{1}{2}a + \frac{1}{2}a^{-1} - 1\). 9. We can multiply both sides of the inequality by \(a + a^{-1}\) to clear the denominator: \[a^n + a^{-n} < \frac{1}{2}(a^n + a^{-n}) + \frac{1}{2}(a + a^{-1}) - 2\] 10. Simplify: \[a^n + a^{-n} < \frac{1}{2}a^n + \frac{1}{2}a^{-n} + \frac{1}{2}a + \frac{1}{2}a^{-1} - 2\] 11. Since \(a > 0\) and \(a \neq 1\), and \(n > 1\), it follows that \(a^n > a\) and \(a^{-n} > a^{-1}\). 12. Therefore, \(a^n + a^{-n} > a + a^{-1}\). 13. Thus, we have proved that \(n^2 < \frac{a^n + a^{-n} - 2}{a + a^{-1} - 2}\).