A triangle $ABC$ is inscribed in a circle $G$. For any point $M$ inside the triangle, $A_1$ denotes the intersection of the ray $AM$ with $G$. Find the locus of point $M$ for which $\frac{BM\cdot CM}{MA_1}$ is minimal, and find this minimum value.
To solve this problem, we need to find the locus of point \( M \) inside triangle \( \triangle ABC \) such that the expression \(\frac{BM \cdot CM}{MA_1}\) is minimized, where \( A_1 \) is the intersection of the ray \( AM \) with the circumcircle \( G \) of \( \triangle ABC \).

1. **Understanding the Geometry**:
   - Let \( \triangle ABC \) be inscribed in a circle \( G \).
   - Let \( M \) be any point inside \( \triangle ABC \).
   - Let \( A_1 \) be the intersection of the ray \( AM \) with the circumcircle \( G \) of \( \triangle ABC \).

2. **Using the Power of a Point Theorem**:
   - By the Power of a Point theorem, we have:
     \[
     AM \cdot MA_1 = BM \cdot MC
     \]
   - This implies:
     \[
     MA_1 = \frac{BM \cdot CM}{AM}
     \]

3. **Expression to Minimize**:
   - We need to minimize the expression:
     \[
     \frac{BM \cdot CM}{MA_1}
     \]
   - Substituting \( MA_1 \) from the Power of a Point theorem, we get:
     \[
     \frac{BM \cdot CM}{\frac{BM \cdot CM}{AM}} = AM
     \]

4. **Minimizing \( AM \)**:
   - To minimize \( AM \), \( M \) should be as close to \( A \) as possible.
   - However, \( M \) must be inside \( \triangle ABC \).

5. **Considering the Incenter**:
   - The incenter \( I \) of \( \triangle ABC \) is equidistant from all sides of the triangle.
   - The incenter \( I \) minimizes the distance to the sides of the triangle, and thus, it is a natural candidate for minimizing the expression.

6. **Verification**:
   - Let \( r \) be the radius of the incircle of \( \triangle ABC \).
   - We need to show that:
     \[
     \frac{BM \cdot CM}{MA_1} \geq 2r
     \]
   - For \( M \) at the incenter \( I \), the expression becomes:
     \[
     \frac{BI \cdot CI}{IA_1}
     \]
   - Since \( I \) is the incenter, \( BI \) and \( CI \) are tangents to the incircle, and \( IA_1 \) is the distance from the incenter to the circumcircle.

7. **Conclusion**:
   - The minimum value of \(\frac{BM \cdot CM}{MA_1}\) is achieved when \( M \) is the incenter \( I \) of \( \triangle ABC \).
   - The minimum value is \( 2r \).

The final answer is \( \boxed{ 2r } \).