Cho a, b, c là các số thực, a, b, c > 0. a, \[ \frac{1}{1 + a^2} + \frac{1}{1 + b^2} > 2 \]

To tackle the problem, let's denote the condition given:

For \( a, b, c > 0 \),
\[
\frac{1}{1 + a^2} + \frac{1}{1 + b^2} > 2
\]

**Step 1: Analyze the inequality.**

This inequality suggests we analyze the function \( f(x) = \frac{1}{1 + x^2} \). We can notice that \( f(x) \) is decreasing for \( x > 0 \). Therefore, for this sum to be greater than 2, \( a \) and \( b \) must be small enough.

If we check with some values:
1. If \( a \) and \( b \) are very small (approaching 0), \( f(a) \) and \( f(b) \) approach 1, thus satisfying \( f(a) + f(b) > 2 \).

This indicates \( a \) and \( b \) cannot increase indefinitely.

**Step 2: Consider \( c \)** and additional conditions.

The next condition states:
\[
\frac{1}{1 + a^3} + \frac{1}{1 + b^3} + \frac{1}{1 + c^3} > 3
\]

Using the same reasoning with \( g(x) = \frac{1}{1 + x^3} \) (which is also decreasing for \( x > 0 \)), we see similar behavior.

**Step 3: Constrain \( a, b, c \)** and their product.

Both conditions highlight that as \( a, b, c \) increase, the corresponding fractions decrease. Therefore, the combination where \( a, b, c \) are close to 0 satisfies both inequalities, but each must remain positive.

This suggests that perhaps small values lead to larger outputs in these functions.

**Conclusion:**

To meet both conditions, \( a, b, c \) should be positive and small enough:
- Specifically, ensuring the conditions hold globally requires \( a, b, c < 1 \) (or even smaller).
- The product \( abc \) would then also ultimately need to remain positive but smaller for both inequalities.

In summary, values of \( a, b, c \) that are sufficiently small will ensure that both inequalities hold true while maintaining positivity.

Further exploration of specific values or additional context from the problem might refine this analysis or provide tighter bounds for potential solutions.