Let $a$ be a real number and $b$ be a real number with $b \neq -1$ and $b \neq 0$. Determine all pairs $(a, b)$ for which

$$
\frac{(1+a)^{2}}{1+b} \leq 1+\frac{a^{2}}{b}
$$

is satisfied. For which pairs $(a, b)$ does equality hold?

(Walther Janous)
The inequality can be equivalently transformed step by step as follows:

\[
\begin{aligned}
& 1+\frac{a^{2}}{b}-\frac{(1+a)^{2}}{1+b} \geq 0 \Longleftrightarrow \\
& \frac{b^{2}+b+a^{2}+a^{2} b-b-2 a b-a^{2} b}{b(1+b)} \geq 0 \Longleftrightarrow \\
& \frac{a^{2}-2 a b+b^{2}}{b(1+b)} \geq 0 \Longleftrightarrow \\
& \frac{(a-b)^{2}}{b(1+b)} \geq 0 .
\end{aligned}
\]

Therefore, equality holds exactly when \( a = b \), i.e., for all pairs \((a, a)\) with \( a \in \mathbb{R} \backslash \{-1, 0\} \).

Now let \( a \neq b \). Then the inequality holds exactly when \( b(1+b) > 0 \), i.e., exactly for \( b > 0 \) or \( b < -1 \), thus for all pairs \((a, b)\) with \( a \in \mathbb{R} \) and \( b \in \mathbb{R} \) such that \( b > 0 \) or \( b < -1 \).

(Walther Janous)