Let $x$ and $y$ be nonnegative real numbers. Prove that $(x +y^3) (x^3 +y) \ge 4x^2y^2$. When does equality holds?

(Task committee)
To prove that \((x + y^3)(x^3 + y) \ge 4x^2y^2\) for nonnegative real numbers \(x\) and \(y\), we will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

1. **Applying AM-GM Inequality:**
   The AM-GM inequality states that for any nonnegative real numbers \(a\) and \(b\),
   \[
   \frac{a + b}{2} \ge \sqrt{ab}
   \]
   with equality if and only if \(a = b\).

2. **Rewriting the Expression:**
   Consider the terms \(x + y^3\) and \(x^3 + y\). We will apply the AM-GM inequality to these terms.

3. **Applying AM-GM to \(x + y^3\):**
   \[
   x + y^3 \ge 2\sqrt{xy^3}
   \]
   This follows from the AM-GM inequality with \(a = x\) and \(b = y^3\).

4. **Applying AM-GM to \(x^3 + y\):**
   \[
   x^3 + y \ge 2\sqrt{x^3y}
   \]
   This follows from the AM-GM inequality with \(a = x^3\) and \(b = y\).

5. **Multiplying the Inequalities:**
   \[
   (x + y^3)(x^3 + y) \ge (2\sqrt{xy^3})(2\sqrt{x^3y})
   \]
   Simplifying the right-hand side:
   \[
   (x + y^3)(x^3 + y) \ge 4\sqrt{xy^3} \cdot \sqrt{x^3y}
   \]
   \[
   (x + y^3)(x^3 + y) \ge 4\sqrt{(xy^3)(x^3y)}
   \]
   \[
   (x + y^3)(x^3 + y) \ge 4\sqrt{x^4y^4}
   \]
   \[
   (x + y^3)(x^3 + y) \ge 4x^2y^2
   \]

6. **Equality Condition:**
   Equality holds in the AM-GM inequality if and only if the terms are equal. Therefore, equality holds if and only if:
   \[
   x = y^3 \quad \text{and} \quad x^3 = y
   \]
   Solving these equations simultaneously, we get:
   \[
   x = y^3 \quad \text{and} \quad (y^3)^3 = y \implies y^9 = y
   \]
   Since \(y\) is nonnegative, the solutions are \(y = 0\) or \(y = 1\). For \(y = 0\), \(x = 0\). For \(y = 1\), \(x = 1\).

Therefore, equality holds when \(x = 0\) and \(y = 0\), or \(x = 1\) and \(y = 1\).

\(\blacksquare\)

The final answer is \( \boxed{ (x + y^3)(x^3 + y) \ge 4x^2y^2 } \) with equality when \(x = 0\) and \(y = 0\), or \(x = 1\) and \(y = 1\).