How do I prove


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

  • $\begingroup$ What range of $x$ and $r$ are you interested in? Is $r \in \mathbb{N}$? $\endgroup$ – Ross Millikan Mar 7 '11 at 18:18
  • $\begingroup$ 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... $\endgroup$ – Fabian Mar 7 '11 at 18:20
  • $\begingroup$ @Fabian, post is edited now, sorry! $\endgroup$ – Milosz Wielondek Mar 7 '11 at 18:30
  • 1
    $\begingroup$ I still get for $r=1$ the expression $1>1$ which is clearly false! $\endgroup$ – Fabian Mar 7 '11 at 18:32
  • $\begingroup$ @Fabian, sorry, I got the domain wrong. It's fixed now! $\endgroup$ – Milosz Wielondek Mar 7 '11 at 18:54

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 $y \gt 0$)

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


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 $r \neq 1$.

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


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.


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