Prove that the volume $V$ and the lateral area $S$ of a right circular cone satisfy the inequality
\[\left( \frac{6V}{\pi}\right)^2 \leq \left( \frac{2S}{\pi \sqrt 3}\right)^3\]
When does equality occur?
1. Let \( r \) be the radius of the base and \( h \) be the vertical height of the right circular cone. The slant height \( l \) is given by \( l = \sqrt{r^2 + h^2} \).

2. The lateral surface area \( S \) of the cone is given by:
   \[
   S = \pi r l = \pi r \sqrt{r^2 + h^2}
   \]

3. The volume \( V \) of the cone is given by:
   \[
   V = \frac{1}{3} \pi r^2 h
   \]

4. We need to prove the inequality:
   \[
   \left( \frac{6V}{\pi} \right)^2 \leq \left( \frac{2S}{\pi \sqrt{3}} \right)^3
   \]

5. Substitute the expressions for \( V \) and \( S \) into the inequality:
   \[
   \left( \frac{6 \cdot \frac{1}{3} \pi r^2 h}{\pi} \right)^2 \leq \left( \frac{2 \cdot \pi r \sqrt{r^2 + h^2}}{\pi \sqrt{3}} \right)^3
   \]

6. Simplify the expressions:
   \[
   \left( 2 r^2 h \right)^2 \leq \left( \frac{2 r \sqrt{r^2 + h^2}}{\sqrt{3}} \right)^3
   \]

7. Further simplification gives:
   \[
   4 r^4 h^2 \leq \left( \frac{2 r \sqrt{r^2 + h^2}}{\sqrt{3}} \right)^3
   \]

8. Simplify the right-hand side:
   \[
   4 r^4 h^2 \leq \frac{8 r^3 (r^2 + h^2)^{3/2}}{3 \sqrt{3}}
   \]

9. Multiply both sides by \( 3 \sqrt{3} \):
   \[
   12 \sqrt{3} r^4 h^2 \leq 8 r^3 (r^2 + h^2)^{3/2}
   \]

10. Divide both sides by \( 4 r^3 \):
    \[
    3 \sqrt{3} r h^2 \leq 2 (r^2 + h^2)^{3/2}
    \]

11. Square both sides to eliminate the square root:
    \[
    (3 \sqrt{3} r h^2)^2 \leq (2 (r^2 + h^2)^{3/2})^2
    \]

12. Simplify:
    \[
    27 r^2 h^4 \leq 4 (r^2 + h^2)^3
    \]

13. Let \( r^2 = a \) and \( h^2 = b \). The inequality becomes:
    \[
    27 a b^2 \leq 4 (a + b)^3
    \]

14. Expand the right-hand side:
    \[
    27 a b^2 \leq 4 (a^3 + 3 a^2 b + 3 a b^2 + b^3)
    \]

15. Simplify:
    \[
    27 a b^2 \leq 4 a^3 + 12 a^2 b + 12 a b^2 + 4 b^3
    \]

16. Rearrange the terms:
    \[
    0 \leq 4 a^3 + 12 a^2 b - 15 a b^2 + 4 b^3
    \]

17. Factorize the expression:
    \[
    0 \leq (2a - b)^2 (a + 4b)
    \]

18. Since \((2a - b)^2 (a + 4b) \geq 0\) is always true, the inequality holds. Equality occurs when \( 2a = b \), i.e., \( 2r^2 = h^2 \). In this case, the slant height \( l \) is equal to \( \sqrt{3} r \).

The final answer is \(\boxed{\left( \frac{6V}{\pi}\right)^2 \leq \left( \frac{2S}{\pi \sqrt 3}\right)^3}\)