Let us consider a convex hexagon ABCDEF. Let $A_1, B_1,C_1, D_1, E_1, F_1$ be midpoints of the sides $AB, BC, CD, DE, EF,FA$ respectively. Denote by $p$ and $p_1$, respectively, the perimeter of the hexagon $ A B C D E F $ and hexagon $ A_1B_1C_1D_1E_1F_1 $. Suppose that all inner angles of hexagon $ A_1B_1C_1D_1E_1F_1 $ are equal. Prove that \[ p \geq \frac{2 \cdot \sqrt{3}}{3} \cdot p_1 .\] When does equality hold ?
1. **Define the Problem and Variables:**
   Let \( A_1, B_1, C_1, D_1, E_1, F_1 \) be the midpoints of the sides \( AB, BC, CD, DE, EF, FA \) respectively of a convex hexagon \( ABCDEF \). Denote by \( p \) the perimeter of hexagon \( ABCDEF \) and by \( p_1 \) the perimeter of hexagon \( A_1B_1C_1D_1E_1F_1 \). We need to prove that:
   \[
   p \geq \frac{2 \cdot \sqrt{3}}{3} \cdot p_1
   \]
   and determine when equality holds.

2. **Introduce Reflections and Angles:**
   Let \( I \) and \( J \) be the reflections of \( B \) and \( C \) through \( A_1B_1 \) and \( B_1C_1 \) respectively. Given that all inner angles of hexagon \( A_1B_1C_1D_1E_1F_1 \) are equal, we have:
   \[
   \angle A_1B_1C_1 = 120^\circ
   \]
   This implies:
   \[
   \angle IB_1J = 60^\circ
   \]

3. **Equilateral Triangle Formation:**
   Since \( IB_1 = BB_1 = CB_1 = JB_1 \), triangle \( \triangle IJB_1 \) is equilateral. Therefore:
   \[
   IJ = IB_1 = BB_1 = b
   \]

4. **Distance Calculation:**
   Consider the distances:
   \[
   a + b + c = A_1I + IJ + JC_1 \geq A_1C_1
   \]
   Using the properties of the equilateral triangle and the Law of Cosines in the context of the hexagon, we get:
   \[
   A_1C_1 = \sqrt{A_1B_1^2 + A_1C_1^2 + A_1B_1 \cdot A_1C_1 \cdot \cos(120^\circ)}
   \]
   Simplifying using \(\cos(120^\circ) = -\frac{1}{2}\):
   \[
   A_1C_1 \geq \frac{(A_1B_1 + A_1C_1)\sqrt{3}}{2} = \frac{(x + y)\sqrt{3}}{2}
   \]

5. **Inequality for Each Side:**
   Similarly, we can derive the inequalities for the other sides:
   \[
   b + c + d \geq \frac{(y + z)\sqrt{3}}{2}
   \]
   \[
   c + d + e \geq \frac{(z + t)\sqrt{3}}{2}
   \]
   \[
   d + e + f \geq \frac{(t + u)\sqrt{3}}{2}
   \]
   \[
   e + f + a \geq \frac{(u + v)\sqrt{3}}{2}
   \]
   \[
   f + a + b \geq \frac{(v + x)\sqrt{3}}{2}
   \]

6. **Summing the Inequalities:**
   Adding all these inequalities, we get:
   \[
   3(a + b + c + d + e + f) \geq (x + y + z + t + u + v)\sqrt{3}
   \]
   Simplifying, we find:
   \[
   a + b + c + d + e + f \geq \frac{x + y + z + t + u + v}{\sqrt{3}}
   \]

7. **Final Inequality:**
   Since \( p = a + b + c + d + e + f \) and \( p_1 = x + y + z + t + u + v \), we have:
   \[
   p \geq \frac{p_1}{\sqrt{3}}
   \]
   Multiplying both sides by \(\frac{2\sqrt{3}}{3}\), we get:
   \[
   p \geq \frac{2\sqrt{3}}{3} \cdot p_1
   \]

8. **Equality Condition:**
   Equality holds when \( ABCDEF \) is a regular hexagon, as all sides and angles are equal, making the midpoints form another regular hexagon.

\(\blacksquare\)

The final answer is \( \boxed{ p \geq \frac{2 \cdot \sqrt{3}}{3} \cdot p_1 } \)