2. Since the sum of any two parentheses on the left side of the original inequality is positive, at most one of the three parentheses is non-positive.
Therefore, we may assume that all three parentheses are positive.
Let $k=abc, a^{3}=\frac{k x}{y}, b^{3}=\frac{k y}{z}, c^{3}=\frac{k z}{x}$.
Then the left side of the original expression
$$=\frac{\left(k^{2} x+z-k y\right)\left(k^{2} y+x-k z\right)\left(k^{2} z+y-k x\right)}{k^{3} x y z} \text {. }$$

Let $k^{2} x+z-k y=u, k^{2} y+x-k z=v$,
$$k^{2} z+y-k x=w \text {. }$$

Then $x=\frac{k u+v}{k^{3}+1}, y=\frac{k v+w}{k^{3}+1}, z=\frac{k w+u}{k^{3}+1}$.
Hence the left side of the original expression
$$=\frac{u v w\left(1+k^{3}\right)^{3}}{k^{3}(k u+v)(k v+w)(k w+u)} \text {. }$$

Notice that,
$$\begin{array}{l} (k u+v)(k v+w)(k w+u) \\ =\left(1+k^{3}\right) u v w+\left(v^{2} w+w^{2} u+u^{2} v\right) k^{2}+ \\ \left(w^{2} v+v^{2} u+u^{2} w\right) k \\ \geqslant\left(1+k^{3}\right) u v w+3 u v w k^{2}+3 u v w k \\ = u v w(k+1)^{3} . \end{array}$$

Then the left side of the original expression
$$\leqslant \frac{\left(1+k^{3}\right)^{3}}{k^{3}(k+1)^{3}}=\left(k+\frac{1}{k}-1\right)^{3} \text {. }$$