8. Positive real numbers $x, y$ satisfy the following conditions: there exist $a \in[0, x], b \in[0, y]$, such that $a^{2}+y^{2}=2, b^{2}+x^{2}=1, a x+b y=1$, then the maximum value of $x+y$ is $\qquad$
Solve
$a x+b y=x \sqrt{2-y^{2}}+y \sqrt{1-x^{2}}=1$, let's assume $x=\sin \alpha, y=\sqrt{2} \sin \beta$, where $\alpha, \beta \in\left[0, \frac{\pi}{2}\right]$. Then we have $\sqrt{2} \sin \alpha \cos \beta+\sqrt{2} \sin \beta \cos \alpha=1 \Rightarrow \sin (\alpha+\beta)=\frac{\sqrt{2}}{2}$. Clearly, $\alpha+\beta=\frac{\pi}{4}$ does not fit the problem, so $\alpha+\beta=\frac{3 \pi}{4}$.
Thus, $x+y=\sin \alpha+\sqrt{2} \sin \beta=\sin \left(\frac{3 \pi}{4}-\beta\right)+\sqrt{2} \sin \beta$ $=\frac{\sqrt{2}}{2} \cos \beta+\frac{3 \sqrt{2}}{2} \sin \beta=\sqrt{5} \sin (\beta+\varphi) \leqslant \sqrt{5}$, equality holds when $x=\frac{2 \sqrt{5}}{5}, y=\frac{3 \sqrt{5}}{5}$. Therefore, the maximum value of $x+y$ is $\sqrt{5}$.