Proof by induction for an exponential inequality: $(2^r-1)(1-x)x^{2^{r}-2}+x^{2^{r}-1}>x^{2^{r}-r-1}$ 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 $r \in \mathbb N$ $(r\neq 1)$?
 A: 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 $r \neq 1$.
A: 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.
A: 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.
