TITLE: Prove Inequality $\frac{\left(1-\alpha\right)\left(1+{\alpha}^{k}\right)}{\left(1+\alpha\right)\left(1-{\alpha}^{k}\right)}\geqslant\frac{1}{k}$
QUESTION [9 upvotes]: I would like to prove the following inequality:
$$ \frac{\left(1 - \alpha \right )\left(1 + {\alpha}^{k} \right )}{\left(1 + \alpha \right )\left(1 - {\alpha}^{k} \right )} \geqslant \frac{1}{k} \ \forall \alpha > 0 ,  k \in \mathbb{N} $$
Any other properties of:
$$ \frac{\left(1 - \alpha \right )\left(1 + {\alpha}^{k} \right )}{\left(1 + \alpha \right )\left(1 - {\alpha}^{k} \right )} $$
Would be great as well.

REPLY [11 votes]: Here's an answer in terms of hyperbolic functions. The (corrected) inequality to be shown is
$${(1 - \alpha)(1 + \alpha^k) \over (1 + \alpha)(1 - \alpha^k)} \geq {1 \over k}$$
This is the same as
$${{(\alpha^{-{1 \over 2}} - \alpha^{{1 \over 2}})(\alpha^{-{k \over 2}} + \alpha^{{k \over 2}})} \over {(\alpha^{-{1 \over 2}} + \alpha^{{1 \over 2}})(\alpha^{-{k \over 2}} - \alpha^{{k \over 2}})}}\geq {1 \over k}$$
For any $\alpha > 0$ there exists some $t$ for which $e^t = \alpha^{1 \over 2}$. The above can be written as 
$${\tanh(t) \over \tanh(kt)} \geq {1 \over k}$$
Since ${\displaystyle {\tanh(t) \over \tanh(kt)}}$ is even, it suffices to prove this for $t > 0$. For $t > 0$ the above equation can be rewritten as 
$$\tanh(kt) \leq k\tanh(t)$$
 The derivative of the left hand side is $k$ sech^2$(kt)$, while the derivative of the right hand side is $k$ sech^2$(t)$. Since sech is decreasing for $t \geq 0$ (this comes from the fact that $\cosh$ is increasing), the left hand side's derivative is less than that of the right. Since the left and right hand sides of this are equal at $t = 0$, the left hand side is always at most the right hand side and the inequality is proven.