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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.

Thank You!

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For k=1 this ratio is -1... – Did Oct 2 '11 at 16:38
You may assume $ k > 1 $. – Drazick Oct 2 '11 at 16:41
Great. For every positive alpha and positive k, the ratio is always negative. – Did Oct 2 '11 at 16:44
@Tony: but the limit at $1$ is $1/k$. – robjohn Oct 2 '11 at 17:41
@robjohn: I didn't say the limit didn't exist. But $0/0$ is undefined, it's as simple as that. – TonyK Oct 2 '11 at 17:59
up vote 6 down vote accepted

For every positive $a\ne1$, define $F(a)$ as $$ F(a)=\frac{(a-1)(a^k+1)}{(a+1)(a^k-1)}. $$ Since $F(a)=F(1/a)$, one can assume that $a>1$. Then, reducing everything to the same denominator, one sees that $F(a)\geqslant1/k$ if and only if $G(a)\geqslant0$, where $$ G(a)=k(a-1)(a^k+1)-(a+1)(a^k-1), $$ that is, $$ G(a) = (k-1) a^{k+1} - (k+1)a^k + (k+1) a -(k-1). $$ This is a polynomial in $a$ and $G(1)=0$. The derivative is $$ G'(a)=(k+1)H(a), \quad H(a)=(k-1) a^{k} - ka^{k-1}+1. $$ One sees that $H(1)=0$ and $$ H'(a)=k(k-1)a^{k-2}(a-1), $$ Hence $H'(a)>0$ for every $a>1$ and $H(a)>H(1)=0$ for every $a>1$. Thus $G$ is increasing on $a\geqslant1$ and we are done.

Edit The same technique can be used to show that the supremum of $F(a)$ is reached when $a\to+\infty$ (and when $a=0$) and that this supremum is $1$. Since $F(a)\to1/k$ when $a\to1$, $a\ne1$, one sees that $1/k<F(a)<1$ for every positive $a\ne1$ and that these inequalities are sharp.

To compute the supremum, assume that $a>1$ and consider the derivative of $\log F$ at $a$. After some simplifications, this has the sign of $$ J(a)=a^{2k}-1-ka^{k-1}(a^2-1). $$ Thus $J(1)=0$ and $J'(a)$ has the sign of $$ K(a)=2a^{k+3}-(k+1)a^2+k-1. $$ Thus $K(1)=0$ and $K'(a)=2(k+3)a^{k+2}-2(k+1)a>2(k+1)a(a^{k+1}-1)>0$. Hence $K$ is increasing on $a>1$, in particular $K(a)>0$. Hence $J$ is increasing on $a>1$, in particular $J(a)>0$. Hence $\log F$ is increasing on $a>1$. Since the limit of $F$ at infinity is $1$, this proves that $F(a)<1$ for every positive $a$, $a\ne1$.

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Didier, I made a mistake as you noticed. I updated the inequality. Sorry for that. You gave the right answer, just make a small modification to the proper notation I have so I could mark you answer. Thank You! – Drazick Oct 2 '11 at 18:11
You mean, replacing $a$ by $\alpha$? What for? – Did Oct 2 '11 at 19:16
I meant the term $ (1 - a) $ instead of $ (a - 1) $. What about the maximum of $ F(a) $. Is it achieved as $ a $ goes to 1? Could it be proved? Thank You. – Drazick Oct 3 '11 at 6:14
See edit. $ $ $ $ – Did Oct 3 '11 at 7:14

You probably want $1-\alpha$ for that first term, otherwise everything is negative. Plotting these for arbitrary $k$, it's easy to see that minimal values occur at $\alpha = 1$. A bit of Mathematica ...

Table[Limit[FullSimplify[( (1 - a) (1 + a^k) ) / ( (1 + a) (1 - a^k) )], a -> 1], {k, 1, 10}]

leads to {1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10}.

Here's a plot of Plot[Table[n ((1 - a) (1 + a^n))/((1 + a) (1 - a^n)), {n, 1, 50}], {a,0,10}]:

Drazick plot

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You were right about the error. I'm looking for an analytic proof. I simulated using MATLAB and got the same result as you got. – Drazick Oct 2 '11 at 18:12

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.

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I couldn't see the move from the second inequality to the third. I'm not familiar with hyperbolic tangent. Looking at Wikipedia didn't make it pop up for me. – Drazick Oct 2 '11 at 18:33
@Drazick ${\displaystyle\tanh(x) = {e^{x} - e^{-x} \over e^{x} + e^{-x}}}$ by definition of the function. Its derivative is $ sech^2(x)$, where ${\displaystyle sech(x) = {2 \over e^x + e^{-x}}}$ – Zarrax Oct 2 '11 at 18:39
@ Zarraz, Yes, I see it now. How could I miss that :-). Thank You. – Drazick Oct 2 '11 at 18:47
What about the maximum of the term on the left of the inequality. Is it achieved as $ a $ goes to 1? Could it be proved? Thank You. – Drazick Oct 3 '11 at 6:15
@Drazick Yes.. $\tanh(kt) = k\tanh(t)$ when $t = 0$ and then $\tanh(kt) < k\tanh(t)$ for $t > 0$ by what I did up there. So ${\displaystyle {\tanh(t) \over \tanh(kt)}}$ starts out at ${1 \over k}$ for $t = 0$ and then is greater than ${1 \over k}$ for $t > 0$ (and for $t < 0$ since the ratio is an even function). Since $e^t = \alpha^{1 \over 2}$, this corresponds to $\alpha = 1$. So this is where the minimum is achieved. – Zarrax Oct 3 '11 at 14:47

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