China Mathematical Competition

In a plane rectangular coordinate system $xOy$, for positive real number pairs $(x, y)$ that satisfy the condition, take points $L(x, 0)$, $M(x, y)$, $N(0, y)$, and then quadrilateral $OLMN$ is a rectangle. Points $P, Q$ are on sides $LM, MN$, respectively, as shown in Fig. 8.1.
Since $a^2 + y^2 = 2$, $b^2 + x^2 = 1$, $ax + by = 1$, we have
$$\begin{aligned} |OP| &= \sqrt{x^2 + b^2} = 1, \\ |OQ| &= \sqrt{a^2 + y^2} = \sqrt{2}, \\ |PQ| &= \sqrt{(a-x)^2 + (b-y)^2} \\ &= \sqrt{(a^2 + y^2) + (b^2 + x^2) - 2(ax + by)} = 1. \end{aligned}$$
Thus, $\triangle OPQ$ is an isosceles right triangle with $P$ as its right-angle vertex.

Fig. 8.1

Therefore, we can set $\angle LOP = \theta$, $\angle QON = \frac{\pi}{4} - \theta$, where $0 \le \theta \le \frac{\pi}{4}$.
Then
$$\begin{aligned} x + y &= |OL| + |ON| \\ &= |OP| \cdot \cos \angle LOP + |OQ| \cdot \cos \angle QON \\ &= \cos \theta + \sqrt{2} \cos \left(\frac{\pi}{4} - \theta\right) = 2 \cos \theta + \sin \theta \\ &= \sqrt{5} \sin(\theta + \varphi), \end{aligned}$$
where $\varphi = \arcsin \frac{2\sqrt{5}}{5}$.
When $\theta = \frac{\pi}{2} - \varphi$ (correspondingly, $x = \frac{2\sqrt{5}}{5}$, $y = \frac{3\sqrt{5}}{5}$), $x + y$ takes the maximum $\sqrt{5}$.

$\square$