Let $a$ and $ b$ be real numbers with $0\le a, b\le 1$. Prove that $$\frac{a}{b + 1}+\frac{b}{a + 1}\le 1$$ When does equality holds?

(K. Czakler, GRG 21, Vienna)
We need to prove that for real numbers \(a\) and \(b\) with \(0 \le a, b \le 1\), the following inequality holds:
\[ \frac{a}{b + 1} + \frac{b}{a + 1} \le 1 \]

1. **Consider the function \( f(a, b) = \frac{a}{b + 1} + \frac{b}{a + 1} \).**
   - We need to show that \( f(a, b) \le 1 \) for \(0 \le a, b \le 1\).

2. **Evaluate \( f(a, b) \) at the boundary values of \(a\) and \(b\).**
   - When \(a = 0\):
     \[
     f(0, b) = \frac{0}{b + 1} + \frac{b}{0 + 1} = b
     \]
     Since \(0 \le b \le 1\), we have \( f(0, b) = b \le 1 \).

   - When \(b = 0\):
     \[
     f(a, 0) = \frac{a}{0 + 1} + \frac{0}{a + 1} = a
     \]
     Since \(0 \le a \le 1\), we have \( f(a, 0) = a \le 1 \).

   - When \(a = 1\):
     \[
     f(1, b) = \frac{1}{b + 1} + \frac{b}{1 + 1} = \frac{1}{b + 1} + \frac{b}{2}
     \]
     We need to show that \( \frac{1}{b + 1} + \frac{b}{2} \le 1 \). Consider the function \( g(b) = \frac{1}{b + 1} + \frac{b}{2} \):
     \[
     g'(b) = -\frac{1}{(b + 1)^2} + \frac{1}{2}
     \]
     Setting \( g'(b) = 0 \):
     \[
     -\frac{1}{(b + 1)^2} + \frac{1}{2} = 0 \implies \frac{1}{(b + 1)^2} = \frac{1}{2} \implies (b + 1)^2 = 2 \implies b + 1 = \sqrt{2} \implies b = \sqrt{2} - 1
     \]
     Since \(0 \le b \le 1\), we check the endpoints:
     \[
     g(0) = 1 + 0 = 1
     \]
     \[
     g(1) = \frac{1}{2} + \frac{1}{2} = 1
     \]
     Thus, \( g(b) \le 1 \) for \(0 \le b \le 1\).

   - When \(b = 1\):
     \[
     f(a, 1) = \frac{a}{1 + 1} + \frac{1}{a + 1} = \frac{a}{2} + \frac{1}{a + 1}
     \]
     We need to show that \( \frac{a}{2} + \frac{1}{a + 1} \le 1 \). Consider the function \( h(a) = \frac{a}{2} + \frac{1}{a + 1} \):
     \[
     h'(a) = \frac{1}{2} - \frac{1}{(a + 1)^2}
     \]
     Setting \( h'(a) = 0 \):
     \[
     \frac{1}{2} - \frac{1}{(a + 1)^2} = 0 \implies \frac{1}{(a + 1)^2} = \frac{1}{2} \implies (a + 1)^2 = 2 \implies a + 1 = \sqrt{2} \implies a = \sqrt{2} - 1
     \]
     Since \(0 \le a \le 1\), we check the endpoints:
     \[
     h(0) = 0 + 1 = 1
     \]
     \[
     h(1) = \frac{1}{2} + \frac{1}{2} = 1
     \]
     Thus, \( h(a) \le 1 \) for \(0 \le a \le 1\).

3. **Combine the results:**
   - We have shown that \( f(a, b) \le 1 \) at the boundary values \(a = 0\), \(b = 0\), \(a = 1\), and \(b = 1\).
   - Since \( f(a, b) \) is continuous and the boundary values satisfy the inequality, by the extreme value theorem, the inequality holds for all \(0 \le a, b \le 1\).

4. **Equality condition:**
   - Equality holds when \(a = 0\), \(b = 0\), \(a = 1\), or \(b = 1\).

The final answer is \(\boxed{ \frac{a}{b + 1} + \frac{b}{a + 1} \le 1 }\)