Suppose $x,y$ are nonnegative real numbers such that $x + y \le 1$. Prove that $8xy \le  5x(1 - x) + 5y(1 - y)$ 
and determine the cases of equality.
1. **Given Condition**: We are given that \(x\) and \(y\) are nonnegative real numbers such that \(x + y \le 1\). We need to prove that:
   \[
   8xy \le 5x(1 - x) + 5y(1 - y)
   \]

2. **Expand the Right-Hand Side**: Let's expand the right-hand side of the inequality:
   \[
   5x(1 - x) + 5y(1 - y) = 5x - 5x^2 + 5y - 5y^2
   \]

3. **Combine Like Terms**: Combine the terms to simplify the expression:
   \[
   5x - 5x^2 + 5y - 5y^2 = 5(x + y) - 5(x^2 + y^2)
   \]

4. **Formulate the Inequality**: We need to show that:
   \[
   8xy \le 5(x + y) - 5(x^2 + y^2)
   \]

5. **Rearrange the Inequality**: Rearrange the inequality to isolate terms involving \(xy\):
   \[
   8xy + 5(x^2 + y^2) \le 5(x + y)
   \]

6. **Use the Given Condition**: Since \(x + y \le 1\), we can substitute \(s = x + y\) and \(p = xy\). Note that \(s \le 1\). The inequality becomes:
   \[
   8p + 5(x^2 + y^2) \le 5s
   \]

7. **Express \(x^2 + y^2\) in Terms of \(s\) and \(p\)**: Using the identity \((x + y)^2 = x^2 + y^2 + 2xy\), we get:
   \[
   s^2 = x^2 + y^2 + 2p \implies x^2 + y^2 = s^2 - 2p
   \]

8. **Substitute \(x^2 + y^2\) Back into the Inequality**: Substitute \(x^2 + y^2 = s^2 - 2p\) into the inequality:
   \[
   8p + 5(s^2 - 2p) \le 5s
   \]

9. **Simplify the Inequality**: Simplify the left-hand side:
   \[
   8p + 5s^2 - 10p \le 5s \implies -2p + 5s^2 \le 5s
   \]

10. **Rearrange and Factor**: Rearrange the inequality:
    \[
    5s^2 - 5s - 2p \le 0
    \]

11. **Isolate \(p\)**: Isolate \(p\) on one side:
    \[
    2p \ge 5s - 5s^2 \implies p \le \frac{5s - 5s^2}{2}
    \]

12. **Check the Maximum Value of \(p\)**: Since \(p = xy\) and by the AM-GM inequality, \(xy \le \left(\frac{x + y}{2}\right)^2\), we have:
    \[
    p \le \left(\frac{s}{2}\right)^2 = \frac{s^2}{4}
    \]

13. **Compare the Two Bounds on \(p\)**: We need to show that:
    \[
    \frac{s^2}{4} \le \frac{5s - 5s^2}{2}
    \]

14. **Simplify the Comparison**: Multiply both sides by 4 to clear the fraction:
    \[
    s^2 \le 10s - 10s^2
    \]

15. **Combine Like Terms**: Combine like terms to form a quadratic inequality:
    \[
    11s^2 \le 10s \implies 11s^2 - 10s \le 0
    \]

16. **Factor the Quadratic**: Factor the quadratic expression:
    \[
    s(11s - 10) \le 0
    \]

17. **Solve the Inequality**: Since \(s \ge 0\) and \(s \le 1\), the inequality holds for \(0 \le s \le \frac{10}{11}\). Given \(s \le 1\), the inequality is satisfied.

18. **Equality Case**: For equality, we need:
    \[
    8xy = 5x(1 - x) + 5y(1 - y)
    \]
    This happens when \(x = y = \frac{1}{2}\), as it satisfies both the given condition \(x + y \le 1\) and the equality condition.

The final answer is \( \boxed{ 8xy \le 5x(1 - x) + 5y(1 - y) } \) with equality when \(x = y = \frac{1}{2}\).