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}
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}
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 $r \neq 1$.
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)
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.