Prove that let $\alpha=a y_{1}+b y_{2}+c y_{3}+d y_{4}, \beta=a x_{4}+b x_{3}+c x_{2}+d x_{1}$, by the Cauchy-Schwarz inequality, we get
$$\begin{aligned} \alpha^{2}= & \left(a y_{1}+b y_{2}+c y_{3}+d y_{4}\right)^{2} \\ \leqslant & {\left[\left(\sqrt{a d} y_{1}\right)^{2}+\left(\sqrt{b c} y_{2}\right)^{2}+\left(\sqrt{b c} y_{3}\right)^{2}+\left(\sqrt{a d} y_{4}\right)^{2}\right] } \\ & {\left[\left(\sqrt{\frac{a}{d}}\right)^{2}+\left(\sqrt{\frac{b}{c}}\right)^{2}+\left(\sqrt{\frac{c}{b}}\right)^{2}+\left(\sqrt{\frac{d}{a}}\right)^{2}\right] } \\ = & \left(a d y_{1}^{2}+b c y_{2}^{2}+b c y_{3}^{2}+a d y_{4}^{2}\right) \cdot\left(\frac{a}{d}+\frac{b}{c}+\frac{c}{b}+\frac{d}{a}\right) \end{aligned}$$

Similarly, we get
$$\beta^{2} \leqslant\left(a d x_{4}^{2}+b c x_{3}^{2}+b c x_{2}^{2}+a d x_{1}^{2}\right) \cdot\left(\frac{a}{d}+\frac{b}{c}+\frac{c}{b}+\frac{d}{a}\right) .$$

Adding them together and using $x_{i}^{2}+y_{i}^{2}=1(i=1,2,3,4), a b+c d=1$, we get
$$\begin{aligned} \alpha^{2}+\beta^{2} & \leqslant(2 a d+2 b c)\left(\frac{a}{d}+\frac{b}{c}+\frac{c}{b}+\frac{d}{a}\right) \\ & =2(a d+b c)\left(\frac{a b+c d}{b d}+\frac{a b+c d}{a c}\right) \\ & =2(a d+b c)\left(\frac{1}{b d}+\frac{1}{a c}\right) \\ & =2\left(\frac{a^{2}+b^{2}}{a b}+\frac{c^{2}+d^{2}}{c d}\right) \end{aligned}$$

Thus, the proposition is proved.