First consider the right inequality. Let $n = p_1^{k_1} p_2^{k_2} \cdots p_s^{k_s}$, (where $p_1, \dots, p_s$ are distinct primes and $k_1, \dots, k_s$ are positive integers). Then
$$\tau(n) = (k_1 + 1)(k_2 + 1) \cdots (k_s + 1)$$
$$\begin{align*} \sigma(n) &= \left(1 + p_1 + \cdots + p_1^{k_1}\right) \left(1 + p_2 + \cdots + p_2^{k_2}\right) \cdots \left(1 + p_s + \cdots + p_s^{k_s}\right) \\ &= \sigma(p_1^{k_1} \cdots p_s^{k_s}). \end{align*}$$
So
$$\frac{\sigma(n)}{\tau(n)\sqrt{n}} = \frac{1+p_1+\cdots+p_1^{k_1}}{(k_1+1)p_1^{\frac{k_1}{2}}} \cdot \frac{1+p_2+\cdots+p_2^{k_2}}{(k_2+1)p_2^{\frac{k_2}{2}}} \cdots \frac{1+p_s+\cdots+p_s^{k_s}}{(k_s+1)p_s^{\frac{k_s}{2}}}.$$
The factors in this product are independent, so it suffices to show that each one is bounded from below by $\frac{3}{2\sqrt{2}}$. We would like to prove that
$$\frac{1 + p + \cdots + p^k}{(k + 1) p^{\frac{k}{2}}} \geq \frac{3}{2\sqrt{2}}$$
for all $p \ge 2$ and $k \ge 1$ (the left-hand side is the smallest when $p = 2, k = 1$).
Let $x = \sqrt{p}$. For every positive integer $k$ the function
$$f_k(x) = \frac{1 + x^2 + x^4 + \dots + x^{2k}}{x^k}$$
is increasing when $x > 1$.
When $k = 1$ and $x > y > 1$ we have
$$f_1(x) - f_1(y) = (x - y) + \frac{y - x}{xy} = (x - y)\left(1 - \frac{1}{xy}\right) > 0.$$
In general
$$f_{2l}(x) = 1 + f_1(x^2) + \cdots + f_1(x^{2l}), \quad (8)$$
$$f_{2l+1}(x) = f_1(x^3) + \cdots + f_1(x^{2l+1}). \tag{9}$$
The functions $f_{2l}(x)$ and $f_{2l+1}(x)$ are obviously increasing. So, $f_k(x)$ is increasing in $x$.
$$\begin{align*} \Rightarrow \frac{1+p+p^2+\cdots+p^k}{p^{\frac{k}{2}}} &\ge \frac{1+2+\cdots+2^k}{2^{\frac{k}{2}}} \\ \Rightarrow \frac{1+p+p^2+\cdots+p^k}{(k+1)p^{\frac{k}{2}}} &\ge \frac{1+2+\cdots+2^k}{(k+1)2^{\frac{k}{2}}} \end{align*}$$
Define $g(k) = \frac{1+2+\cdots+2^k}{(k+1)2^{\frac{k}{2}}}$, $k \in \mathbb{N}$. Let us show that $g(k)$ is increasing.
$$\begin{align*} g(k+1) - g(k) &= \frac{1 + 2(1 + 2 + \cdots + 2^k)}{(k+2)2^{\frac{k}{2}}\sqrt{2}} - \frac{1 + 2 + \cdots + 2^k}{(k+1)2^{\frac{k}{2}}} \\ &= \frac{(k+1) + (1 + 2 + \cdots + 2^k)(2k + 2 - \sqrt{2}k - 2\sqrt{2})}{(k+1)(k+2)2^{\frac{k}{2}}\sqrt{2}} > 0 \end{align*}$$
So,
$$\frac{1+p+p^2+\dots+p^k}{(k+1)p^{\frac{k}{2}}} \ge \frac{1+2+\dots+2^k}{(k+1)2^{\frac{k}{2}}} \ge \frac{3}{2\sqrt{2}}$$
We conclude that
$$\frac{\sigma(n)}{\tau(n)\sqrt{n}} \ge \left( \frac{3}{2\sqrt{2}} \right)^s \ge \frac{3}{2\sqrt{2}}$$
The equality holds when $n = 2$.

Finally, consider the left inequality. Denote by $1 = d_1 < d_2 < \dots < d_{\tau(n)} = n$ all divisors of $n$. Then $\sigma(n) = \sum_{i=1}^{\tau(n)} d_i$. Since $\frac{n}{d_1}, \frac{n}{d_2}, \dots, \frac{n}{d_{\tau(n)}}$ are precisely the divisors of $n$, we also have $\sigma(n) = \sum_{i=1}^{\tau(n)} \frac{n}{d_i}$.

For each divisor $d_i$ we also have
$$d_i + \frac{n}{d_i} \le n + 1 \iff d_i^2 - d_i(n + 1) + n \le 0 \iff (n - d_i)(1 - d_i) \le 0, \quad (10)$$
and the equality holds if and only if either $d_i = 1$ or $d_i = n$. Now, we can estimate
$$\begin{aligned} \frac{\sigma(n)}{\tau(n)} &= \frac{\sigma(n) + \sigma(n)}{2\tau(n)} = \frac{\sum_{i=1}^{\tau(n)} d_i + \sum_{i=1}^{\tau(n)} \frac{n}{d_i}}{2\tau(n)} = \frac{\sum_{i=1}^{\tau(n)} \left(d_i + \frac{n}{d_i}\right)}{2\tau(n)} \le \\ &\le \frac{\sum_{i=1}^{\tau(n)} (n+1)}{2\tau(n)} = \frac{\tau(n) \cdot (n+1)}{2\tau(n)} = \frac{n+1}{2}. \end{aligned}$$
The equality holds if and only if it holds for each (10). This is only the case when $\tau(n) \le 2$. So, the equality holds when either $n = 1$ or $n$ is a prime.