I will answer a question slightly different from yours. Let $a(n)$ be the smallest integer $\geq n$ such that $a(n) = b c$ with $(b,c) = 1$ and $b \geq n^r$, $c \geq n^{1-r}$. Then :

- $a(n) - n \geq \frac{1}{2} n^{1-r}$ for infinitely many $n$.
- $a(n) - n \ll n^{1-r} \log \log(3n)$.

The first point simply follows from the observation that the fractional part of $n^r$ is in $]0,\frac{1}{2}[$ for infinitely many $n$, so that $b \geq n^{r} + \frac{1}{2}$ in the decomposition $a(n) = bc$. Thus
$a(n) = bc \geq (n^{r} + \frac{1}{2})n^{1-r} = n + \frac{1}{2} n^{1-r}. $

The second point follows from standard point-counting methods, by using the inequality
$$\mathbb{1}_{(b,c)=1} \geq 1 - \sum_p \mathbb{1}_{p|b,p|c} .$$

**Added :** Let me expand what I meant by "standard point-counting methods". One has, for $\delta > n^{-r}$,
$$ \sum_{b > n^{r}, c > n^{1-r} \\ bc \leq n(1+ \delta)} \mathbb{1}_{(b,c)=1} \geq N(n^r,n^{1-r},\delta) - \sum_p N \left( \frac{n^r}{p},\frac{n^{1-r}}{p}, \delta \right),$$
where
$$
N(X,Y,\delta) = \sum_{b > X, c > Y \\ bc \leq XY(1+ \delta)} 1.
$$
Now, let's assume $\delta < 1$. Then $N(X,Y,\delta) = 0$ if $X < (1+ \delta)^{-1}$ or $Y < (1+ \delta)^{-1}$. So let's assume that $X,Y \geq (1+ \delta)^{-1}$ (corresponding to $p \leq (1+\delta) n^r)$ above). Now,
$$
N(X,Y,\delta) = \sum_{X < b \leq X(1 + \delta)} \left( \sum_{Y < c \leq \frac{XY(1+\delta)}{b}} 1\right) \\ = \sum_{X < b \leq X(1 + \delta)} \left( \frac{XY(1+\delta)}{b} - Y + O(1) \right) \\ = f(\delta) XY + O(1 + \delta X + \delta Y),
$$
where $f(\delta) = (1 + \delta) \log(1+\delta) - \delta$. This implies
$$N(n^r,n^{1-r},\delta) - \sum_{p \leq \delta n^{\frac{1}{2}}} N \left( \frac{n^r}{p},\frac{n^{1-r}}{p}, \delta \right) \geq c f(\delta) n + O(\delta n^{1-r} \log \log(3n)),$$
where $c = 1 - \sum_{p} p^{-2}$ is a positive constant. For $p > \delta n^{\frac{1}{2}}$, the quantity $N \left( \frac{n^r}{p},\frac{n^{1-r}}{p}, \delta \right)$ counts at most one $b$, and there are $\ll n^{\frac{1}{2} - r}$ choices for $c$, hence
$$
N \left( \frac{n^r}{p},\frac{n^{1-r}}{p}, \delta \right) \ll n^{\frac{1}{2} - r} \sum_{n^r<bp\leq n^r(1+\delta)} 1.
$$
Setting $k = pb$, we get
$$
\sum_{p > \delta n^{\frac{1}{2}}} N \left( \frac{n^r}{p},\frac{n^{1-r}}{p}, \delta \right) \ll n^{\frac{1}{2} - r} \sum_{n^r<k\leq n^r(1+\delta)} \omega(k) \ll \delta n^{\frac{1}{2}} \log(3n)
$$
Using that $f(\delta) \asymp \delta^2$, one gets the result with $\delta = Cn^{-r} \log \log(3n)$, the constant $C$ being chosen large enough.