Proof that $\left(2-\frac 1r\right)x^r+1\leq (x+1)^r$ for $x\in (0,1)$ and $r\in (1,2)$ Let $x\in (0,1)\ \&\ r\in (1,2)$.
How to show that $$\left(2-\frac 1r\right)x^r+1\leq (x+1)^r$$

Let $f(x) = (x+1)^r - \left(2 - \frac{1}{r}\right)x^r$ where $x$ and $r$ satisfies your conditions.
We have $f^{'}(x) = r(x+1)^{r-1} -  \left(2 - \frac{1}{r}\right)rx^{r-1}$
We have $f^{'}(x) = 0 \leftrightarrow r(x+1)^{r-1} = \left(2 - \frac{1}{r}\right)rx^{r-1}$
Therefore $x = \frac{1}{\left( 2-\frac{1}{r}\right)^{\frac{1}{r-1}} - 1}.$
Define $x_0:= \frac{1}{\left( 2-\frac{1}{r}\right)^{\frac{1}{r-1}} - 1}$ and note that $f^{'}(x) >0$  if $x > x_0$ and $f^{'}(x) >0$  if $x < x_0$. Thus $x_0 $is the global minimum of $f.$
A direct computation leads to
$$f(x_0)= \frac{\left(  2 - \frac{1}{r}\right)^{\frac{r}{r-1}} - \left(  2 - \frac{1}{r}\right)}{\left(\left(  2 - \frac{1}{r}\right) - 1\right)^r}$$
Since $2-\frac{1}{r} >1$ and $\frac{r}{r-1}> 1$ we have $f(x_0)>0$
Therefore 
$$ \left(  2 - \frac{1}{r}\right)x^r \leq (x+1)^r $$
I think that you only need to consider $x>0$ instead of $x \in (0,1)$
