Question 71: 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 $\qquad$ _.
Question 71, Solution: According to the Fibonacci identity: $\left(a^{2}+y^{2}\right)\left(b^{2}+x^{2}\right)=(ax+by)^{2}+(xy-ab)^{2}$, and $a \leq x, b \leq y$, hence $xy-ab=1$.

Let $a=\sqrt{2} \cos \alpha, y=\sqrt{2} \sin \alpha, x=\cos \beta, b=\sin \beta$, then $\alpha, \beta \in\left[0, \frac{\pi}{2}\right]$. Since $1=xy-ab=\sqrt{2} \sin \alpha \cdot \cos \beta-\sqrt{2} \cos \alpha \cdot \sin \beta=\sqrt{2} \sin (\alpha-\beta)$, it follows that $\alpha-\beta=\frac{\pi}{4}$. Therefore,
$$
x+y=\sqrt{2} \sin \alpha+\cos \beta=\sqrt{2} \sin \left(\beta+\frac{\pi}{4}\right)+\cos \beta=\sin \beta+2 \cos \beta \leq \sqrt{5}
$$

When $\cos \beta=\frac{2 \sqrt{5}}{5}, \sin \beta=\frac{\sqrt{5}}{5}$, i.e., $x=\frac{3 \sqrt{5}}{5}, y=\frac{2 \sqrt{5}}{5}$, the equality holds.
Note: On the other hand, according to the above analysis, $\beta \in\left[0, \frac{\pi}{4}\right]$, hence $x+y=\sin \beta+2 \cos \beta \geq 2$, when $\cos \beta=1, \sin \beta=0$, i.e., $x=1, y=1$, the equality holds.