If $ a$ and $ b$ are nonnegative real numbers with $ a^2\plus{}b^2\equal{}4$, show that: $ \frac{ab}{a\plus{}b\plus{}2} \le \sqrt{2}\minus{}1$ and determine when equality occurs.
1. Given the condition \(a^2 + b^2 = 4\), we need to show that:
   \[
   \frac{ab}{a+b+2} \le \sqrt{2} - 1
   \]
   and determine when equality occurs.

2. Let's start by rewriting the expression \(\frac{ab}{a+b+2}\). We will use the fact that \(a^2 + b^2 = 4\).

3. First, we express \(a\) and \(b\) in terms of their sum and product. Let \(s = a + b\) and \(p = ab\). Then, we have:
   \[
   a^2 + b^2 = s^2 - 2p
   \]
   Given \(a^2 + b^2 = 4\), we substitute:
   \[
   s^2 - 2p = 4 \implies p = \frac{s^2 - 4}{2}
   \]

4. Substitute \(p\) into the original inequality:
   \[
   \frac{ab}{a+b+2} = \frac{p}{s+2} = \frac{\frac{s^2 - 4}{2}}{s+2} = \frac{s^2 - 4}{2(s+2)}
   \]

5. Simplify the expression:
   \[
   \frac{s^2 - 4}{2(s+2)} = \frac{(s-2)(s+2)}{2(s+2)} = \frac{s-2}{2}
   \]

6. We need to show that:
   \[
   \frac{s-2}{2} \le \sqrt{2} - 1
   \]

7. Multiply both sides by 2:
   \[
   s - 2 \le 2\sqrt{2} - 2
   \]

8. Add 2 to both sides:
   \[
   s \le 2\sqrt{2}
   \]

9. Since \(s = a + b\) and \(a^2 + b^2 = 4\), we use the Cauchy-Schwarz inequality:
   \[
   (a + b)^2 \le (1^2 + 1^2)(a^2 + b^2) = 2 \cdot 4 = 8
   \]
   Therefore:
   \[
   s^2 \le 8 \implies s \le \sqrt{8} = 2\sqrt{2}
   \]

10. Hence, we have shown that:
    \[
    s \le 2\sqrt{2}
    \]
    which implies:
    \[
    \frac{s-2}{2} \le \sqrt{2} - 1
    \]

11. Equality holds when \(s = 2\sqrt{2}\). For \(a\) and \(b\) to satisfy \(a + b = 2\sqrt{2}\) and \(a^2 + b^2 = 4\), we solve the system:
    \[
    a + b = 2\sqrt{2}
    \]
    \[
    a^2 + b^2 = 4
    \]
    This system has the solution \(a = b = \sqrt{2}\), which gives:
    \[
    ab = (\sqrt{2})(\sqrt{2}) = 2
    \]

The final answer is \( \boxed{ \frac{ab}{a+b+2} \le \sqrt{2} - 1 } \) and equality holds when \(a = b = \sqrt{2}\).