To prove the inequality

\[
\frac{1}{a^3 + b^3 + 1} + \frac{1}{b^3 + c^3 + 1} + \frac{1}{c^3 + a^3 + 1} < 1,
\]

given that \(a, b, c > 0\) and \(abc = 1\), we can use the method of substitution and the Cauchy-Schwarz inequality.

**Step 1: Using the identity \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\)**

By applying the AM-GM inequality, we know:

\[
a^3 + b^3 \geq ab(a + b).
\]

**Step 2: Apply Cauchy-Schwarz inequality**

Using Cauchy-Schwarz in the following way:

\[
\left( \frac{1}{a^3 + b^3 + 1} + \frac{1}{b^3 + c^3 + 1} + \frac{1}{c^3 + a^3 + 1} \right) \left( (a^3 + b^3 + 1) + (b^3 + c^3 + 1) + (c^3 + a^3 + 1) \right) \geq (1 + 1 + 1)^2 = 9.
\]

**Step 3: Simplify the second term**

The term

\[
(a^3 + b^3 + 1) + (b^3 + c^3 + 1) + (c^3 + a^3 + 1) = 2(a^3 + b^3 + c^3) + 3.
\]

**Step 4: Establishing a bound**

Considering more closely \(a^3 + b^3 + c^3\):

Using the relation from the identity,

\[
a^3 + b^3 + c^3 \geq 3\sqrt[3]{(abc)^3} = 3,
\]

we have:

\[
2(a^3 + b^3 + c^3) + 3 \geq 2(3) + 3 = 9.
\]

**Step 5: Combining inequalities**

From the Cauchy-Schwarz result:

\[
\left( \frac{1}{a^3 + b^3 + 1} + \frac{1}{b^3 + c^3 + 1} + \frac{1}{c^3 + a^3 + 1} \right) \cdot (2(a^3 + b^3 + c^3) + 3) \geq 9,
\]

Let \( S = \frac{1}{a^3 + b^3 + 1} + \frac{1}{b^3 + c^3 + 1} + \frac{1}{c^3 + a^3 + 1} \), then

\[
S \cdot (2(a^3 + b^3 + c^3) + 3) \geq 9.
\]

Thus, if we let \(2(a^3 + b^3 + c^3) + 3 \geq 9\), it simplifies:

\[
S < 1.
\]

Therefore, we conclude:

\[
\frac{1}{a^3 + b^3 + 1} + \frac{1}{b^3 + c^3 + 1} + \frac{1}{c^3 + a^3 + 1} < 1.
\]

This proves the desired inequality.