Since the inequality involves the minimum of two positive numbers and since the function $y = x^2$ is increasing on the set $\mathbb{R}^+$, our task is to verify
$$(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \le \left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)^2 \quad \text{and} \quad (x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \le \left(\frac{y}{x}+\frac{x}{z}+\frac{z}{y}\right)^2, \quad (1)$$

as well as to find when *at least one* equality in (1) holds. Replacing a triple ($x$, $y$, $z$)
by the triple ($y$, $x$, $z$), we get the second inequality in (1) from the first one. Thus we
can restrict to the proof of the first inequality. Distributing both sides leads to
$$3 + \frac{x}{y} + \frac{x}{z} + \frac{y}{x} + \frac{y}{z} + \frac{z}{x} + \frac{z}{y} \le \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} + 2\left(\frac{x}{z} + \frac{y}{x} + \frac{z}{y}\right).$$
Let us introduce the new (positive) variables $a = x/y$, $b = y/z$, $c = z/x$ and rewrite
the last inequality as
$$\left(a^2 - 1 - a + \frac{1}{a}\right) + \left(b^2 - 1 - b + \frac{1}{b}\right) + \left(c^2 - 1 - c + \frac{1}{c}\right) \ge 0. \quad (2)$$
For any positive $t$, we notice that
$$t^2 - 1 - t + \frac{1}{t} = (t^2 - 1) - \frac{t^2 - 1}{t} = \frac{(t^2 - 1)(t - 1)}{t} = \frac{(t - 1)^2(t + 1)}{t}.$$
This implies that (2) holds as well and that (2) becomes an equality if and only if $a = b = c = 1$, i.e. $x = y = z$ for the original variables. Note that the last condition does not change under transformation $(x, y, z) \to (y, x, z)$. Thus the original inequality is proven and becomes an equality if and only $x = y = z$.