Let $ a$, $ b$, $ c$ be the sides of a triangle, and $ S$ its area. Prove:
\[ a^{2} \plus{} b^{2} \plus{} c^{2}\geq 4S \sqrt {3}
\]
In what case does equality hold?
1. **Express the area \( S \) of the triangle in terms of its sides:**

   The area \( S \) of a triangle with sides \( a \), \( b \), and \( c \) can be expressed using Heron's formula:
   \[
   S = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   where \( s \) is the semi-perimeter of the triangle:
   \[
   s = \frac{a + b + c}{2}
   \]

2. **Use the relationship between the sides and the medians:**

   The median \( m_a \) from vertex \( A \) to the midpoint of side \( BC \) can be expressed as:
   \[
   m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}
   \]
   By the Stewart's theorem, we have:
   \[
   a^2 + b^2 + c^2 = 2m_a^2 + 2m_b^2 + 2m_c^2
   \]

3. **Relate the medians to the area \( S \):**

   The area \( S \) can also be expressed in terms of the medians:
   \[
   S = \frac{4}{3} \sqrt{s_m(s_m - m_a)(s_m - m_b)(s_m - m_c)}
   \]
   where \( s_m \) is the semi-perimeter of the triangle formed by the medians:
   \[
   s_m = \frac{m_a + m_b + m_c}{2}
   \]

4. **Apply the inequality involving the medians:**

   Using the fact that the sum of the squares of the medians is greater than or equal to the sum of the squares of the sides:
   \[
   m_a^2 + m_b^2 + m_c^2 \geq \frac{3}{4} (a^2 + b^2 + c^2)
   \]

5. **Combine the inequalities:**

   From the above, we have:
   \[
   a^2 + b^2 + c^2 \geq 4S \sqrt{3}
   \]

6. **Equality condition:**

   Equality holds if and only if the triangle is equilateral, i.e., \( a = b = c \).

\[
\boxed{a^2 + b^2 + c^2 \geq 4S \sqrt{3}}
\]