Solution:
Let
$$\begin{aligned} & S_{a} = a + a b + a b c \\ & S_{b} = b + b c + b c d \\ & S_{c} = c + c d + c d a \\ & S_{d} = d + d a + d a b. \end{aligned}$$
Notice that $1 + S_{a} = \frac{1}{2} + \left(\frac{1}{2} + S_{a}\right) \geq 2 \sqrt{\frac{1}{2} \cdot \left(\frac{1}{2} + S_{a}\right)} = \sqrt{2} \cdot \sqrt{\frac{1}{2} + S_{a}}$. Using similar relations for $S_{b}$, $S_{c}$, and $S_{d}$ we see that the left-hand side of the required inequality is greater than or equal to $D = \sqrt{2}\left(\frac{1}{1 + S_{a}} + \frac{1}{1 + S_{b}} + \frac{1}{1 + S_{c}} + \frac{1}{1 + S_{d}}\right)$. We now have $1 + S_{a} = a + a b + a b c + a b c d = a \cdot (1 + S_{b}) = a b \cdot (1 + S_{c}) = a b c (1 + S_{d})$. Likewise, $1 + S_{b} = b c (1 + S_{d})$ and $1 + S_{c} = c (1 + S_{d})$, which yields $D = \sqrt{2} \cdot \frac{1}{1 + S_{d}}\left(\frac{1}{a b c} + \frac{1}{b c} + \frac{1}{c} + 1\right) = \sqrt{2}$. Thus the statement is proved.

Choose positive $w, x, y, z$ such that
$$a = \frac{x}{w}, \quad b = \frac{y}{x}, \quad c = \frac{z}{y}, \quad d = \frac{w}{z}$$
Then we have
$$\frac{1}{2} + a + a b + a b c = \frac{1}{2} + \frac{x}{w} + \frac{y}{w} + \frac{z}{w} = \frac{1}{2} + \frac{x + y + z}{w} = \frac{1}{2} + \frac{\frac{1}{2} - w}{w} = \frac{1}{2} \cdot \frac{1 - w}{w}$$
and similarly for the other three terms. Thus each term of the sum has the form $\sqrt{\frac{2 w}{1 - w}}$.

If $0 < x < \frac{1}{2}$, then $\sqrt{\frac{w}{1 - w}} > 2 w$

$$\sqrt{\frac{x}{1 - x}} > 2 x$$
Square both sides and multiply by $1 - x$ to get
$$x > 4 x^{2} (1 - x)$$
Dividing by $x$ and simplifying,
$$0 > 4 x (1 - x) - 1 = -4 x^{2} + 4 x - 1$$
This last expression is $-(2 x - 1)^{2}$ which is negative since $x < \frac{1}{2}$. (Note that all the steps of this are reversible, so this final true inequality can be used to work backwards and establish our desired inequality.)

Using this result, the original expression is greater than (never equal to)
$$\sqrt{2} \cdot (2 w + 2 x + 2 y + 2 z) = \sqrt{2}$$