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

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 start by applying the method of Titu's lemma (also known as the Cauchy-Schwarz inequality).

Using the Cauchy-Schwarz inequality, we can rewrite each fraction:

\[
\frac{1}{a^3 + b^3 + 1} < \frac{1}{a^3 + b^3 + c^3} \quad \text{(since } c^3 > 0\text{)},
\]
\[
\frac{1}{b^3 + c^3 + 1} < \frac{1}{b^3 + c^3 + a^3},
\]
\[
\frac{1}{c^3 + a^3 + 1} < \frac{1}{c^3 + a^3 + b^3}.
\]

Now summing these inequalities gives us:

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

Next, we can apply the AM-GM inequality:

\[
a^3 + b^3 + c^3 \geq 3\sqrt[3]{(abc)^3} = 3\sqrt[3]{1} = 3.
\]

Thus, we have:

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

Now, we note that:

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

which implies that:

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

Consequently, we have proved the required inequality:

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