# 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)?

Sep 10, 2016

Let a > 1. Then

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

$\left({a}^{- n} / {a}^{- 1}\right) \left(\frac{{a}^{2 n} - 2 {a}^{n} + 1}{{a}^{2} - 2 a + 1}\right)$

$= \frac{1}{a} ^ \left(n - 1\right) {\left(\frac{{a}^{n} - 1}{a - 1}\right)}^{2}$

$= \frac{1}{a} ^ \left(n - 1\right) {\left(1 + a + {a}^{2} + {a}^{3} + \ldots + {a}^{n - 1}\right)}^{2}$

$> {\left(1 + 1 + 1 + \ldots . + 1\right)}^{2} / {a}^{n - 1}$

, using ${a}^{r} > 1 , r = 1. 2. 3 , \ldots$

Likewise, when a < 1,

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

$\left({a}^{n} / a\right) \left(\frac{{a}^{- 2 n} - 2 {a}^{- n} + 1}{{a}^{- 2} - 2 {a}^{- 1} + 1}\right)$

$= {a}^{n - 1} {\left(\frac{1 - {a}^{- \left(n - 1\right)}}{1 - {a}^{- 1}}\right)}^{2}$

$= {a}^{n - 1} {\left(1 + \frac{1}{a} + \frac{1}{a} ^ 2 + \frac{1}{a} ^ 3 + \ldots + \frac{1}{a} ^ \left(n - 1\right)\right)}^{2}$

$> {a}^{n - 1} {\left(1 + 1 + 1 + \ldots . + 1\right)}^{2}$

>n^2a^(n-1

, using ${a}^{- r} > 1 , r = 1. 2. 3 , \ldots$

So, the given expression is

$> {n}^{2} / {a}^{n - 1}$, for $a > 1$ and

$> {n}^{2} {a}^{n - 1}$, for a < 1#.

Sep 10, 2016

${a}^{n} + {a}^{- n} - 2 = {\left({a}^{\frac{n}{2}} - {a}^{- \frac{n}{2}}\right)}^{2}$

then

$\frac{{a}^{n} + {a}^{- n} - 2}{a + {a}^{- 1} - 2} = {\left(\frac{{b}^{n} - {b}^{- n}}{b - {b}^{- 1}}\right)}^{2}$

with $b = \sqrt{a}$ so

$n < \frac{{b}^{n} - {b}^{- n}}{b - {b}^{- 1}}$ calling now $b = {e}^{\lambda}$ we have

$n < \frac{{\left({e}^{\lambda}\right)}^{n} - {\left({e}^{\lambda}\right)}^{- n}}{{e}^{\lambda} - {\left({e}^{\lambda}\right)}^{- 1}} = \frac{{e}^{\lambda n} - {e}^{- \lambda n}}{{e}^{\lambda} - {e}^{- \lambda}} = \sinh \frac{\lambda n}{\sinh} \left(\lambda\right)$

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

also we have

${\lim}_{\lambda \to 0} \left(\sinh \frac{\lambda n}{\sinh} \left(\lambda\right)\right) = n$

so the proof follows.