All sides of a convex hexagon $ABCDEF$ are $1$. Let $M,m$ be the maximum and minimum possible values of three diagonals $AD,BE,CF$. Find the range of $M$, $m$.
1. **Understanding the Problem:**
   We are given a convex hexagon \(ABCDEF\) with all sides equal to 1. We need to find the maximum (\(M\)) and minimum (\(m\)) possible values of the lengths of the diagonals \(AD\), \(BE\), and \(CF\).

2. **Maximum Value \(M\):**
   - Consider the diagonal \(AD\). By the triangle inequality, the length of \(AD\) must be less than or equal to the sum of the lengths of the sides of the hexagon that connect \(A\) to \(D\):
     \[
     |AD| \leq |AB| + |BC| + |CD| = 1 + 1 + 1 = 3
     \]
   - This reasoning applies similarly to the diagonals \(BE\) and \(CF\). Therefore, the maximum possible value \(M\) for any of these diagonals is:
     \[
     M = 3
     \]

3. **Minimum Value \(m\):**
   - Since the hexagon is convex, all internal angles are less than \(180^\circ\). This implies that the shortest distance between any two non-adjacent vertices (i.e., the diagonals) must be at least 1.
   - To see why, consider the diagonal \(AD\). Since \(A\) and \(D\) are not adjacent, the shortest path between them in a convex hexagon is a straight line, which must be at least 1 unit long.
   - Therefore, the minimum possible value \(m\) for any of these diagonals is:
     \[
     m = 1
     \]

4. **Conclusion:**
   - The range of the possible values for the diagonals \(AD\), \(BE\), and \(CF\) is from 1 to 3.

The final answer is \( \boxed{ [1, 3] } \)