The answer is \(1 + 2 \sqrt{2}\).

If we expand the expression under the square root on the left, we get:

\[2 \cdot \sqrt{3 + \frac{x}{y} + \frac{x}{z} + \frac{y}{x} + \frac{y}{z} + \frac{z}{x} + \frac{z}{y}} - \sqrt{\left(1 + \frac{x}{y} \right) \left(1 + \frac{y}{z} \right)}\]

By setting \(a = x/y\) and \(b = y/z\), we can simplify the expression to only have two variables:

\[f(a, b) = 2 \cdot \sqrt{3 + a + a \cdot b + \frac{1}{a} + b + \frac{1}{a \cdot b} + \frac{1}{b}} - \sqrt{(1 + a) (1 + b)}\]

Here, due to symmetry between \(a\) and \(b\) (i.e., \(f(a,b) = f(b,a)\)), it seems plausible that the minimum occurs at \(a = b\). However, this assumption is not necessarily correct. For example, function \(g(a,b) = [(a-1)^2+(b-2)^2]\cdot[(a-2)^2+(b-1)^2]\) is also symmetric, but it minimizes at \((1,2)\) and \((2,1)\). To assume that the minimum occurs at \(a=b\), we must examine other properties of the function.

Thus, we do not assume this and continue by simplifying \(f\) as follows:

\[f(a, b) = 2 \cdot \sqrt{\left(1 + a + \frac{1}{b}\right)\left(1 + b + \frac{1}{a}\right)} - \sqrt{(1 + a) (1 + b)}\]

The Cauchy-Schwarz inequality states that for any two arbitrary vectors \(u\) and \(v\), the absolute value of their inner product is bounded by the product of their norms:

\[|u \cdot v| \leq ||u|| \cdot ||v||\]

If we choose:

\[\begin{aligned} u &= \left(\frac{1}{\sqrt{b}}, \sqrt{1+a}\right) \\\\ v &= \left(\frac{1}{\sqrt{a}}, \sqrt{1+b}\right) \end{aligned}\]

then their inner product and norms are:

\[\begin{aligned} u \cdot v &= \frac{1}{\sqrt{b}} \cdot \frac{1}{\sqrt{a}} + \sqrt{1+a} \cdot \sqrt{1+b} = \frac{1}{\sqrt{a \cdot b}} + \sqrt{(1 + a)(1 + b)} \\\\ ||u|| &= \sqrt{\left( \frac{1}{\sqrt{b}} \right)^2 + \left( \sqrt{1+a} \right)^2} = \sqrt{1 + a + \frac{1}{b}} \\\\ ||v|| &= \sqrt{\left(\frac{1}{\sqrt{a}}\right)^2 + \left(\sqrt{1+b}\right)^2} = \sqrt{1 + b + \frac{1}{a}}. \end{aligned}\]

Thus, by the Cauchy-Schwarz inequality we have:

\[\label{inequality:first} \frac{1}{\sqrt{a \cdot b}} + \sqrt{(1 + a)(1 + b)} \leq \sqrt{\left(1 + a + \frac{1}{b}\right)\left(1 + b + \frac{1}{a}\right)}\]

Now, if we define function \(e\) as:

\[\begin{aligned} e(a, b) &= 2 \cdot \left(\frac{1}{\sqrt{a \cdot b}} + \sqrt{(1+a)(1+b)} \right) - \sqrt{(1+a)(1+b)} \\\\ &= \frac{2}{\sqrt{a \cdot b}} + \sqrt{(1+a)(1+b)}. \end{aligned}\]

Then, according to Inequality [inequality:first], we have \(f(a,b) \geq e(a, b)\).

Also, by the AM-GM inequality for any two nonnegative numbers \(x\) and \(y\) we have:

\[\frac{x+y}{2} \geq \sqrt{x\cdot y}.\]

Using this, we get:

\[\label{inequality:second} (1+a)(1+b) = 1 + a \cdot b + a + b \geq 1 + a \cdot b + 2\sqrt{a \cdot b} = (1 + \sqrt{a \cdot b})^2.\]

Now, defining function \(d\) as:

\[d(a, b) = \frac{2}{\sqrt{a \cdot b}} + 1 + \sqrt{a \cdot b}.\]

By Inequality [inequality:second], we get \(e(a, b) \geq d(a, b)\), and thus \(f(a, b) \geq d(a, b)\). Therefore, the minimum value of \(f\) is greater than or equal to the minimum of the much simpler function \(d\).

By setting \(t = \sqrt{a \cdot b}\) and analyzing the derivatives of \(d\), we find its minimum at \(\sqrt{a \cdot b} = t = \sqrt{2}\), which equals \(1 + 2 \sqrt{2}\). Also, we have \(f(\sqrt{2}, \sqrt{2}) = 1 + 2 \sqrt{2}\), and since the minimum value of \(f\) is greater than or equal to the minimum of \(d\), the minimum of \(f\) is also \(1 + 2 \sqrt{2}\).