$M$ is an arbitrary point inside $\triangle ABC$. $AM$ intersects the circumcircle of the triangle again at $A_1$. Find the points $M$ that minimise $\frac{MB\cdot MC}{MA_1}$.
1. **Using the Power of a Point Theorem:**
   The Power of a Point theorem states that for a point \( M \) inside a circle, the power of \( M \) with respect to the circle is given by:
   \[
   MA \cdot MA_1 = MB \cdot MC
   \]
   where \( A_1 \) is the second intersection of line \( AM \) with the circumcircle of \( \triangle ABC \).

2. **Expression to Minimize:**
   We need to minimize the expression:
   \[
   \frac{MB \cdot MC}{MA_1}
   \]
   Using the Power of a Point theorem, we can rewrite this as:
   \[
   \frac{MB \cdot MC}{MA_1} = \frac{MB \cdot MC}{\frac{MB \cdot MC}{MA}} = MA
   \]
   Therefore, we need to minimize \( MA \).

3. **Using the Pedal Triangle:**
   Let \( A'B'C' \) be the pedal triangle of \( \triangle ABC \) with respect to point \( M \). The sides of \( \triangle A'B'C' \) are given by the projections of \( M \) onto the sides of \( \triangle ABC \). By the Law of Sines, the lengths of the sides of \( \triangle A'B'C' \) are:
   \[
   A'B' = AM \sin A, \quad B'C' = BM \sin B, \quad C'A' = CM \sin C
   \]

4. **Area of the Pedal Triangle:**
   The area \( K_{A'B'C'} \) of the pedal triangle can be expressed as:
   \[
   K_{A'B'C'} = \frac{R^2 - OM^2}{4R^2} \cdot K_{ABC}
   \]
   where \( R \) is the circumradius of \( \triangle ABC \) and \( OM \) is the distance from the circumcenter \( O \) to the point \( M \).

5. **Relating the Areas:**
   The area of \( \triangle A'B'C' \) can also be expressed in terms of the sides:
   \[
   4R_{A'B'C'} K_{A'B'C'} = AM \cdot BM \cdot CM \cdot \frac{abc}{8R^3}
   \]
   where \( R_{A'B'C'} \) is the circumradius of \( \triangle A'B'C' \).

6. **Minimizing the Expression:**
   We need to minimize:
   \[
   \frac{AM \cdot BM \cdot CM}{R^2 - OM^2}
   \]
   Using the relationship between the areas, we have:
   \[
   \frac{AM \cdot BM \cdot CM}{R^2 - OM^2} = 2R_{A'B'C'}
   \]
   Since \( R_{A'B'C'} \geq r \) (the inradius of \( \triangle ABC \)), the minimum value occurs when \( R_{A'B'C'} = r \).

7. **Conclusion:**
   The minimum value of the expression \( \frac{MB \cdot MC}{MA_1} \) is achieved when \( M \) is the incenter \( I \) of \( \triangle ABC \).

\[
\boxed{2r}
\]