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.

How do I prove

$(2^r-1)(1-x)x^{2^{r}-2}+x^{2^{r}-1}>x^{2^{r}-r-1}$

for $\frac{1}{2}<x<1$ and $r \in \mathbb N$ $(r\neq 1)$?

share|improve this question
    
What range of $x$ and $r$ are you interested in? Is $r \in \mathbb{N}$? –  Ross Millikan Mar 7 '11 at 18:18
    
For what $r$ and what $x$ does this inequality supposedly hold. For $r=1$ it is always wrong. For $r=0$ it is wrong for $x\geq1$ and $x<0$. This where the only two cases I checked so far... –  Fabian Mar 7 '11 at 18:20
    
@Fabian, post is edited now, sorry! –  Milosz Wielondek Mar 7 '11 at 18:30
1  
I still get for $r=1$ the expression $1>1$ which is clearly false! –  Fabian Mar 7 '11 at 18:32
    
@Fabian, sorry, I got the domain wrong. It's fixed now! –  Milosz Wielondek Mar 7 '11 at 18:54

3 Answers 3

up vote 5 down vote accepted

Assuming $1/2 \lt x \lt 1$ and $r \in \mathbb{N}$,

Set $y = 1/x$ and multiply by $y^{2^{r} - 1}$ to get (which is equivalent to the original for $y \gt 0$)

$$(2^r - 1)(y-1) + 1 \gt y^r$$

Now

Which is true if (as $y \gt 1$)

$$(2^r - 1) \gt \frac{y^r-1}{y-1}$$

Which is same as

$$ 1 + 2 + 2^2 + \dots + 2^{r-1} \gt 1 + y + y^2 + \dots + y^{r-1}$$

Which is true if $ 1 \lt y \lt 2$ (i.e $1/2 \lt x \lt 1$) and $r \neq 1$.

share|improve this answer
    
Yes, you are completely right. I messed up the domain, it is as you stated it $\frac{1}{2}<x<1$. –  Milosz Wielondek Mar 7 '11 at 18:53
    
Do you think you could edit your answer to match the new domain? –  Milosz Wielondek Mar 7 '11 at 19:59
    
@Milosz: Done.. –  Aryabhata Mar 7 '11 at 20:40
    
Thanks! How did you simplify $\frac{y^{r}-1}{y-1}$ into $1+y+y^{2}+\ldots+y^{r-1}$? –  Milosz Wielondek Mar 11 '11 at 3:55
    
@Milosz: That is a geometric progression. –  Aryabhata Mar 11 '11 at 5:12

The assertion holds for every $x$ in $(1/2,1)$ and every real number $r>1$.

To see this, one can begin like Moron and try to show that $f(y)<f(2)$ with $y=1/x$ and $$ f(y)=(y^r-1)/(y-1). $$ Since $y<2$, it is enough to prove that $f$ is nondecreasing. The sign of $f'(y)$ is the sign of $g(y)$ with $$ g(y)=(r-1)y^r-ry^{r-1}+1. $$ Now, $g'(y)=r(r-1)y^{r-2}(y-1)$ is positive on $y>1$ because $r>1$ hence $y\mapsto g(y)$ is increasing on $y>1$, and $g(1)=0$ hence $g(y)>0$ for $y>1$. Finally, $y\mapsto f(y)$ is increasing on $y>1$ hence we are done.

share|improve this answer

The inequality is equivalent to $(2^r-1)(1-x)x^{r-1}+x^r>1$, that is $(2^r-1)x^{r-1}-(2^r-2)x^r>1$. To prove this one you can differentiate the LHS with respect to $x$, to find $x^{r-2} \left((2^r-1)(r-1)-(2^r-2)r x \right)$. From that you get that the expression $(2^r-1)(1-x)x^{r-1}+x^r$ is either (as a function of $x$) increasing on $[1/2,1]$, increasing on $[1/2,x_0]$ and decreasing on $[x_0,1]$, or decreasing on $[1/2,1]$ (actually it is the second one since $x_0 = \frac{(2^r-1)(r-1)}{(2^r-2)r}$ is between $1/2$ and $1$, but it doesn't matter). So we only need to check that the values at $x=1/2$ and $x=1$ are $\geq 1$, and in fact they are $1$, so that the inequality is sharp.

share|improve this answer

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.