9・14 For all $x, y$ that satisfy the inequality
$$
\log _{x^{2}+y^{2}}(x+y) \geqslant 1
$$

find the maximum value of $y$.
［Solution］
$$
\begin{array}{l}
\text { Let } M=\left\{(x, y) ; x+y \geqslant x^{2}+y^{2}>1\right\}, \\
N=\left\{(x, y) ; 0<x+y \leqslant x^{2}+y^{2}<1\right\} .
\end{array}
$$

It is clear that the solution set of the inequality $\log _{x^{2}+y^{2}}(x+y) \geqslant 1$ is $M \cup N$.
If $(x, y) \in N$, then $y<1$. If $(x, y) \in M$, then by
$$
\begin{array}{l}
\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2} \text { we know that } \\
y \leqslant \frac{1}{2}+\frac{\sqrt{2}}{2} .
\end{array}
$$

On the other hand, it is clear that $\left(\frac{1}{2}, \frac{1}{2}+\frac{\sqrt{2}}{2}\right) \in M$. Therefore, $y=\frac{1}{2}+\frac{\sqrt{2}}{2}$ is the maximum value of $y$ sought.