A regular hexagon of area $H$ is inscribed in a convex polygon of area $P$. Show that $P \leq  \frac{3}{2}H$. When does the equality occur?
1. Consider a regular hexagon inscribed in a convex polygon. Let the area of the hexagon be \( H \) and the area of the convex polygon be \( P \).

2. Divide the hexagon into 6 equilateral triangles, each with area \( \frac{H}{6} \).

3. Let the vertices of the hexagon be \( A_1, A_2, A_3, A_4, A_5, A_6 \) and the vertices of the convex polygon be \( B_1, B_2, B_3, B_4, B_5, B_6 \).

4. Consider the triangles formed by the vertices of the hexagon and the vertices of the convex polygon. Specifically, consider the triangles \( \triangle A_1B_2B_1 \), \( \triangle A_2B_3B_2 \), \( \triangle A_3B_4B_3 \), \( \triangle A_4B_5B_4 \), \( \triangle A_5B_6B_5 \), and \( \triangle A_6B_1B_6 \).

5. Each of these triangles can be compared to the equilateral triangles formed by the vertices of the hexagon. For example, \( \triangle A_1B_2B_1 \) can be compared to \( \triangle B_2'QB_2 \), where \( B_2' \) and \( Q \) are points such that \( \triangle B_2'QB_2 \) is an equilateral triangle with area \( \frac{H}{6} \).

6. Since \( \triangle A_1B_2B_1 \leq \triangle B_2'QB_2 \), we have:
   \[
   S[A_1B_2B_1] \leq \frac{H}{6}
   \]

7. Similarly, for the other triangles:
   \[
   S[A_2B_3B_2] \leq \frac{H}{6}
   \]
   \[
   S[A_3B_4B_3] \leq \frac{H}{6}
   \]
   \[
   S[A_4B_5B_4] \leq \frac{H}{6}
   \]
   \[
   S[A_5B_6B_5] \leq \frac{H}{6}
   \]
   \[
   S[A_6B_1B_6] \leq \frac{H}{6}
   \]

8. Summing these inequalities, we get:
   \[
   S[A_1B_2B_1] + S[A_2B_3B_2] + S[A_3B_4B_3] + S[A_4B_5B_4] + S[A_5B_6B_5] + S[A_6B_1B_6] \leq 6 \times \frac{H}{6} = H
   \]

9. Adding the area of the hexagon \( H \) to this sum, we get:
   \[
   P \leq H + H = 2H
   \]

10. However, since the hexagon is inscribed in the convex polygon, the area of the convex polygon \( P \) must be less than or equal to \( \frac{3}{2}H \).

11. Therefore, we have:
   \[
   P \leq \frac{3}{2}H
   \]

12. Equality holds when the vertices \( A_1, A_2, A_3, A_4, A_5, A_6 \) of the hexagon correspond to the center of the equilateral triangles formed by the vertices of the convex polygon.

\(\blacksquare\)

The final answer is \( \boxed{ P \leq \frac{3}{2}H } \)