Let the angles of a triangle be $\alpha, \beta$, and $\gamma$, the perimeter $2p$ and the radius of the circumcircle $R$. Prove the inequality $\cot^2 \alpha + \cot^2 \beta + \cot^2 \gamma \ge  3 \left(\frac{9R^2}{p^2}-1\right)$. When is the equality achieved?
1. **Expressing cotangent in terms of sides and circumradius:**
   Recall that for a triangle with angles $\alpha, \beta, \gamma$ and sides $a, b, c$ opposite these angles respectively, the cotangent of an angle can be expressed using the sides and the circumradius $R$:
   \[
   \cot \alpha = \frac{b^2 + c^2 - a^2}{4R \cdot \text{area of the triangle}}
   \]
   Similarly, we can write expressions for $\cot \beta$ and $\cot \gamma$.

2. **Using the identity for the sum of cotangents squared:**
   We know that:
   \[
   \cot^2 \alpha + \cot^2 \beta + \cot^2 \gamma = \frac{(b^2 + c^2 - a^2)^2}{16R^2 \cdot (\text{area})^2} + \frac{(a^2 + c^2 - b^2)^2}{16R^2 \cdot (\text{area})^2} + \frac{(a^2 + b^2 - c^2)^2}{16R^2 \cdot (\text{area})^2}
   \]

3. **Simplifying the expression:**
   Using the fact that the area of the triangle can be expressed as:
   \[
   \text{area} = \frac{abc}{4R}
   \]
   we can substitute this into the cotangent expressions. However, a more straightforward approach is to use known identities and inequalities involving the cotangent functions.

4. **Using known inequalities:**
   We use the identity:
   \[
   \cot^2 \alpha + \cot^2 \beta + \cot^2 \gamma = \frac{a^2 + b^2 + c^2}{4R^2} - 1
   \]
   and the fact that:
   \[
   a + b + c = 2p
   \]
   Therefore:
   \[
   a^2 + b^2 + c^2 \geq \frac{(a + b + c)^2}{3} = \frac{(2p)^2}{3} = \frac{4p^2}{3}
   \]

5. **Substituting and simplifying:**
   \[
   \cot^2 \alpha + \cot^2 \beta + \cot^2 \gamma = \frac{a^2 + b^2 + c^2}{4R^2} - 1 \geq \frac{\frac{4p^2}{3}}{4R^2} - 1 = \frac{p^2}{3R^2} - 1
   \]
   We need to show:
   \[
   \frac{p^2}{3R^2} - 1 \geq 3 \left(\frac{9R^2}{p^2} - 1\right)
   \]

6. **Rewriting the inequality:**
   \[
   \frac{p^2}{3R^2} - 1 \geq 3 \left(\frac{9R^2}{p^2} - 1\right)
   \]
   Simplifying the right-hand side:
   \[
   3 \left(\frac{9R^2}{p^2} - 1\right) = \frac{27R^2}{p^2} - 3
   \]
   Therefore, we need to show:
   \[
   \frac{p^2}{3R^2} - 1 \geq \frac{27R^2}{p^2} - 3
   \]

7. **Cross-multiplying and simplifying:**
   \[
   \frac{p^4}{3R^2} - p^2 \geq 27R^2 - 3p^2
   \]
   \[
   \frac{p^4}{3R^2} + 2p^2 \geq 27R^2
   \]
   Multiplying through by $3R^2$:
   \[
   p^4 + 6p^2R^2 \geq 81R^4
   \]

8. **Equality condition:**
   The equality holds when the triangle is equilateral, i.e., $\alpha = \beta = \gamma = \frac{\pi}{3}$.

\[
\boxed{\cot^2 \alpha + \cot^2 \beta + \cot^2 \gamma \ge  3 \left(\frac{9R^2}{p^2}-1\right)}
\]