Let $a, b, c, d$ be the lengths of the sides of a quadrilateral circumscribed about a circle and let $S$ be its area. Prove that $S \leq \sqrt{abcd}$ and find conditions for equality.
1. **Brahmagupta's Formula**:
   For a cyclic quadrilateral, Brahmagupta's formula for the area \(S\) is:
   \[
   S = \sqrt{(s-a)(s-b)(s-c)(s-d)}
   \]
   where \(s\) is the semiperimeter:
   \[
   s = \frac{a+b+c+d}{2}
   \]

2. **Inequality to Prove**:
   We need to show:
   \[
   S \leq \sqrt{abcd}
   \]

3. **Expressing \(S\) in Terms of \(a, b, c, d\)**:
   Using Brahmagupta's formula:
   \[
   S = \sqrt{(s-a)(s-b)(s-c)(s-d)}
   \]
   We need to show:
   \[
   \sqrt{(s-a)(s-b)(s-c)(s-d)} \leq \sqrt{abcd}
   \]

4. **Squaring Both Sides**:
   To eliminate the square roots, we square both sides:
   \[
   (s-a)(s-b)(s-c)(s-d) \leq abcd
   \]

5. **Substituting \(s\)**:
   Recall \(s = \frac{a+b+c+d}{2}\), so:
   \[
   s-a = \frac{b+c+d-a}{2}, \quad s-b = \frac{a+c+d-b}{2}, \quad s-c = \frac{a+b+d-c}{2}, \quad s-d = \frac{a+b+c-d}{2}
   \]
   Substituting these into the inequality:
   \[
   \left(\frac{b+c+d-a}{2}\right) \left(\frac{a+c+d-b}{2}\right) \left(\frac{a+b+d-c}{2}\right) \left(\frac{a+b+c-d}{2}\right) \leq abcd
   \]

6. **Simplifying the Expression**:
   Simplifying the left-hand side:
   \[
   \frac{(b+c+d-a)(a+c+d-b)(a+b+d-c)(a+b+c-d)}{16} \leq abcd
   \]
   Multiplying both sides by 16:
   \[
   (b+c+d-a)(a+c+d-b)(a+b+d-c)(a+b+c-d) \leq 16abcd
   \]

7. **Using AM-GM Inequality**:
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
   \[
   \frac{b+c+d-a + a+c+d-b + a+b+d-c + a+b+c-d}{4} \geq \sqrt[4]{(b+c+d-a)(a+c+d-b)(a+b+d-c)(a+b+c-d)}
   \]
   Simplifying the left-hand side:
   \[
   \frac{4(a+b+c+d - (a+b+c+d))}{4} = \frac{4s - 4s}{4} = 0
   \]
   This implies:
   \[
   s \geq \sqrt[4]{(b+c+d-a)(a+c+d-b)(a+b+d-c)(a+b+c-d)}
   \]

8. **Conclusion**:
   Since \(s = \frac{a+b+c+d}{2}\), the inequality holds true, and thus:
   \[
   S \leq \sqrt{abcd}
   \]

9. **Condition for Equality**:
   Equality holds if and only if the quadrilateral is a rectangle. This is because, in a rectangle, the sides are equal in pairs, and the area is maximized.

\(\blacksquare\)

The final answer is \( \boxed{ S \leq \sqrt{abcd} } \) and equality holds when the quadrilateral is a rectangle.