We say that a polygon $P$ is [i]inscribed[/i] in another polygon $Q$ when all vertices of $P$ belong to perimeter of $Q$. We also say in this case that $Q$ is [i]circumscribed[/i] to $P$. Given a triangle $T$, let $l$ be the maximum value of the side of a square inscribed in $T$ and $L$ be the minimum value of the side of a square circumscribed to $T$. Prove that for every triangle $T$ the inequality $L/l \ge 2$ holds and find all the triangles $T$ for which the equality occurs.
To prove that for every triangle \( T \), the inequality \( \frac{L}{l} \ge 2 \) holds, and to find all triangles \( T \) for which the equality occurs, we need to analyze the maximum side length \( l \) of a square inscribed in \( T \) and the minimum side length \( L \) of a square circumscribed around \( T \).

1. **Expression for \( l \):**
   The side length \( l \) of the largest square inscribed in a triangle \( T \) can be expressed as:
   \[
   l = \max \left( \frac{a h_a}{a + h_a}, \frac{b h_b}{b + h_b}, \frac{c h_c}{c + h_c} \right)
   \]
   where \( a, b, c \) are the sides of the triangle, and \( h_a, h_b, h_c \) are the corresponding altitudes.

2. **Expression for \( L \):**
   The side length \( L \) of the smallest square circumscribed around a triangle \( T \) can be expressed as:
   \[
   L = \min \left( \max(a, h_a), \max(b, h_b), \max(c, h_c) \right)
   \]

3. **Case Analysis:**
   We will analyze different cases to show that \( \frac{L}{l} \ge 2 \).

   **Case 1:** \( a = \min \left( \max(a, h_a), \max(b, h_b), \max(c, h_c) \right) \)

   - **Subcase 1.1:** \( a + h_a = \min (a + h_a, b + h_b, c + h_c) \)
     \[
     \frac{L}{l} = \frac{a}{\frac{a h_a}{a + h_a}} = \frac{a (a + h_a)}{a h_a} = \frac{a + h_a}{h_a} = \frac{a}{h_a} + 1 \ge 1 + 1 = 2
     \]

   - **Subcase 1.2:** \( b + h_b = \min (a + h_a, b + h_b, c + h_c) \)
     - If \( a \ge h_a \) and \( b \ge h_b \), then \( a \le b \) implies \( h_a \ge h_b \). Thus, \( b \ge a \ge h_a \ge h_b \). Since \( a h_a = b h_b \), we get \( a = b \) and \( h_a = h_b \), reducing to Subcase 1.1.
     - If \( h_b \ge b \), then \( a \le h_b \) and \( a \ge h_a \). Thus, \( h_b \ge a \ge h_a \) and \( h_a \ge b \). Since \( a h_a = b h_b \), we get \( h_b \ge a \ge h_a \ge b \), reducing to \( a = h_b \) and \( b = h_a \), which is equivalent to Subcase 1.1.

   **Case 2:** \( h_a = \min \left( \max(a, h_a), \max(b, h_b), \max(c, h_c) \right) \)

   - **Subcase 2.1:** \( a + h_a = \min (a + h_a, b + h_b, c + h_c) \)
     \[
     \frac{L}{l} = \frac{h_a}{\frac{a h_a}{a + h_a}} = \frac{h_a (a + h_a)}{a h_a} = \frac{a + h_a}{a} = \frac{h_a}{a} + 1 \ge 1 + 1 = 2
     \]

   - **Subcase 2.2:** \( b + h_b = \min (a + h_a, b + h_b, c + h_c) \)
     - If \( b \ge h_b \), then \( h_a \le b \). Since \( a h_a = b h_b \), we get \( b \ge h_a \ge a \ge h_b \). But \( b + h_b \ge a + h_a \), reducing to \( b = h_a \), which is equivalent to Subcase 2.1.
     - If \( b \le h_b \), then \( h_a \le h_b \) and \( a \ge b \). Thus, \( h_b \ge h_a \ge a \ge b \), reducing to \( a = b \), which is equivalent to Subcase 2.1.

4. **Equality Condition:**
   Equality holds when \( a = h_a \) or \( b = h_b \) or \( c = h_c \). This occurs when the triangle is a right triangle with the right angle at one of its vertices.

\(\blacksquare\)

The final answer is \( \boxed{ \frac{L}{l} \ge 2 } \) and equality holds for right triangles with the right angle at one of its vertices.