Prove that $a^{\frac{2n}{m}} - 1 \geq n\big(a^{\frac{n+1}m} - a^{\frac{n-1}{m}}\big)$. 
Given natural numbers $m,n,$ and a real number $a>1$, prove the inequality :
$$\displaystyle a^{\frac{2n}{m}} - 1 \geq n\big(a^{\frac{n+1}m} - a^{\frac{n-1}{m}}\big)$$
SOURCE : Inequalities (PDF) (Page Number 2 ; Question Number 153.2)

I have been trying this problem from 2 weeks but still no success. I tried every method I could think of like AM-GM, C-S, Holder and more, but could not find a proof.
Also, is it necessary for $n,m$ to be natural numbers ?
Any help will be gratefully acknowledged.
Thanks in advance ! :)
 A: The answer can be given via simple calculus, and thus the result can be shown to hold true for all $x$. However, the OP has stated that he/she would prefer a solution that did not resort to calculus. So here is my edited answer. For my original answer, please check the edit history. 
Let $a^{\frac{1}{m}}=x>1$. The question is equivalent to showing $$x^{2n}-1 \ge n(x^{n+1}-x^{n-1}) \iff \frac{x^{2n}-1}{x^2-1} \ge nx^{n-1}$$
Now, note that $$\frac{x^{2n}-1}{x^2-1}=\sum_{k=0}^{n-1}x^{2k}=\frac{1}{2} \left(\sum_{k=0}^{n-1}x^{2k}+x^{2n-2k-2}\right) \ge \frac{1}{2} \times 2\sum_{k=0}^{n-1}x^{n-1}=nx^{n-1}$$
From $\text{AM-GM}$. Our proof is done. 
A: Let $x = a^{\frac 1m} > 1$. Using
$$
x^{2n} - 1 = (x-1)(1+ x+x^2 + \ldots + x^{2n-1}) \\
x^{n+1} - x^{n-1} = (x-1) (x^{n-1}+x^n)
$$
we get
$$
x^{2n} - 1 - n(x^{n+1} - x^{n-1}) = 
(x-1)\left( 1+ x+x^2 + \ldots + x^{2n-1} - n(x^{n-1}+x^n) \right) \\
= (x-1)\sum_{k=1}^n \left( x^{k-1} + x^{2n-k} - x^{n-1}-x^n\right) \\
 = (x-1)\sum_{k=1}^n (x^{n-k}-1)(x^n - x^{k-1}) \\ \ge 0
$$
(with strict inequality for $n \ge 2$), which is the desired
inequality
$$
 x^{2n} - 1 \ge n(x^{n+1} - x^{n-1}) \, .
$$
This proof works for positive real $m$ and integer $n \ge 1$.
For $x < 1$ the same inequality with $\ge$ replaced by $\le$
holds.
