Minimum value of $i$ to change $\lfloor n / i \rfloor$

I was solving this CSES question called Sum of Divisors and one of the solutions in the USACO Guide hints at this statement:

The minimum value of $j > i$ such that $\lfloor n/j \rfloor$ < $\lfloor n/i \rfloor$ is $j = \lfloor n/q \rfloor + 1$, where $q = \lfloor n/i \rfloor$ for integers $j, i \le n$.

I am trying to prove this.

Let $q = \lfloor n/i \rfloor$, then since $q \le \dfrac ni$, we have $\dfrac nq \ge i$. But since $i$ is an integer we can also write $$\left\lfloor \dfrac nq \right\rfloor \ge i \implies 1 + \left\lfloor \dfrac nq \right\rfloor \ge i + 1$$ and so $$\dfrac{n}{1 + \left\lfloor \dfrac nq \right\rfloor} \le \frac{n}{i+1}\lt\frac ni\implies \left\lfloor\dfrac{n}{1 + \left\lfloor \dfrac nq \right\rfloor}\right\rfloor \le \left\lfloor\frac ni\right\rfloor$$ $$\implies \left\lfloor\frac nj\right\rfloor \le \left\lfloor\frac ni\right\rfloor$$

1. I need to prove that $\left\lfloor\dfrac nj\right\rfloor \lt \left\lfloor\dfrac ni\right\rfloor$. How do I get the strict inequality?
2. I also need to prove that $\left\lfloor\dfrac nq\right\rfloor + 1$ is the minimum such $j$ possible. For that we would want $\lfloor n / \lfloor n/q\rfloor\rfloor = q$, for which I give a partial attempt again:

$$q = \left\lfloor \dfrac ni \right\rfloor \implies qi \le n \lt (q + 1)i$$, then $$i \le \frac nq \lt \frac {q+1}{q}i\implies i \le \left\lfloor \dfrac nq \right\rfloor\lt i + \left\lfloor \dfrac iq \right\rfloor$$ Inverting this and trying to get $\lfloor n / \lfloor n/q\rfloor\rfloor$ just seems too clunky after this.