Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
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
1  
@Tony: but the limit at $1$ is $1/k$. –  robjohn Oct 2 '11 at 17:41
2  
@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
show 3 more comments

3 Answers

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 $a$, $a\ne1$.

share|improve this answer
    
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
1  
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
add comment

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.

share|improve this answer
    
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
add comment

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.