Does this inequality hold for $r>1$? Does this inequality hold?

Let $r>1$ and $i<j$ where $i$ and $j$ are positive integers, then $$\frac{1-\big(\frac{1}{r^j}\big)}{r^\frac{j}{2}-\big(\frac{1}{r^j}\big)}<\frac{1-\big(\frac{1}{r^j}\big)}{r^\frac{i}{2}-\big(\frac{1}{r^j}\big)}<\frac{1-\big(\frac{1}{r^i}\big)}{r^\frac{i}{2}-\big(\frac{1}{r^i}\big)}$$ 

Actually, I get $$\frac{1-\big(\frac{1}{r^j}\big)}{r^\frac{j}{2}-\big(\frac{1}{r^j}\big)}<\frac{1-\big(\frac{1}{r^j}\big)}{r^\frac{i}{2}-\big(\frac{1}{r^j}\big)}$$  but can't understand why $$\frac{1-\big(\frac{1}{r^j}\big)}{r^\frac{i}{2}-\big(\frac{1}{r^j}\big)}<\frac{1-\big(\frac{1}{r^i}\big)}{r^\frac{i}{2}-\big(\frac{1}{r^i}\big)}$$
Please help. Thank you.

Your second inequality seems to be false in general. Indeed, it is equivalent to
$$\frac{r^j-1}{r^{\frac{i}{2}+j}-1} < \frac{r^i-1}{r^{\frac{3i}{2}}-1} \tag{1}$$
And since both denominators are positive, this is equivalent to
$$
(r^j-1)(r^{\frac{3i}{2}}-1) < (r^i-1)(r^{\frac{i}{2}+j}-1) \tag{2}
$$
Expanding, 
$$
r^{j+\frac{3i}{2}}-r^j-r^{\frac{3i}{2}}+1 < r^{j+\frac{3i}{2}}-r^i-r^{\frac{i}{2}+j}+1 \tag{3}
$$
which is equivalent to
$$
-r^j-r^{\frac{3i}{2}} < -r^i-r^{\frac{i}{2}+j}\tag{4}
$$
Or
$$
r^{\frac{i}{2}+j}-r^j < r^{\frac{3i}{2}}-r^i\tag{5}
$$
which is equivalent to
$$
r^j < \frac{r^{\frac{3i}{2}}-r^i}{r^{\frac{i}{2}}-1}=r^i \tag{6}
$$
This is false as $i < j$.
