Find the largest value of the expression  $\frac{p}{R}\left( 1- \frac{r}{3R}\right)$ , where $p,R, r$ is, respectively, the perimeter, the radius of the circumscribed circle and the radius of the inscribed circle of a triangle.
To find the largest value of the expression $\frac{p}{R}\left( 1- \frac{r}{3R}\right)$, where $p$, $R$, and $r$ are the perimeter, the radius of the circumscribed circle, and the radius of the inscribed circle of a triangle respectively, we will proceed as follows:

1. **Express the given expression in terms of known trigonometric identities:**

   \[
   \frac{p}{R}\left( 1- \frac{r}{3R}\right)
   \]

2. **Use the known relationships in a triangle:**

   - The perimeter $p$ of a triangle is given by $p = a + b + c$.
   - The radius $R$ of the circumscribed circle is given by $R = \frac{abc}{4K}$, where $K$ is the area of the triangle.
   - The radius $r$ of the inscribed circle is given by $r = \frac{K}{s}$, where $s = \frac{p}{2}$ is the semi-perimeter of the triangle.

3. **Express $p$ in terms of $R$ and the angles of the triangle:**

   \[
   p = 2R(\sin A + \sin B + \sin C)
   \]

   Therefore,

   \[
   \frac{p}{R} = 2(\sin A + \sin B + \sin C)
   \]

4. **Substitute this into the original expression:**

   \[
   \frac{p}{R}\left( 1- \frac{r}{3R}\right) = 2(\sin A + \sin B + \sin C) \left( 1- \frac{r}{3R}\right)
   \]

5. **Simplify the expression:**

   \[
   2(\sin A + \sin B + \sin C) \left( 1- \frac{r}{3R}\right) = 2(\sin A + \sin B + \sin C) - \frac{2r(\sin A + \sin B + \sin C)}{3R}
   \]

6. **Use the known inequality for the sum of sines in a triangle:**

   \[
   \sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2}
   \]

   This inequality achieves equality when the triangle is equilateral.

7. **Substitute the maximum value of $\sin A + \sin B + \sin C$:**

   \[
   2 \left( \frac{3\sqrt{3}}{2} \right) - \frac{2r \left( \frac{3\sqrt{3}}{2} \right)}{3R} = 3\sqrt{3} - \frac{\sqrt{3}r}{R}
   \]

8. **Use the relationship between $r$ and $R$ in an equilateral triangle:**

   In an equilateral triangle, $r = \frac{R}{2}$.

   \[
   3\sqrt{3} - \frac{\sqrt{3} \cdot \frac{R}{2}}{R} = 3\sqrt{3} - \frac{\sqrt{3}}{2}
   \]

9. **Simplify the final expression:**

   \[
   3\sqrt{3} - \frac{\sqrt{3}}{2} = \frac{6\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}
   \]

Therefore, the largest value of the expression is:

\[
\boxed{\frac{5\sqrt{3}}{2}}
\]