Proof I Set $u = ay_1 + by_2$, $v = cy_3 + dy_4$, $u_1 = ax_4 + bx_3$ and $v_1 = cx_2 + dx_1$. Then
$$u^2 \le (ay_1 + by_2)^2 + (ax_1 - bx_2)^2 \\ = a^2 + b^2 + 2ab(y_1 y_2 - x_1 x_2),$$
that is
$$x_1 x_2 - y_1 y_2 \le \frac{a^2 + b^2 - u^2}{2ab}. \qquad ①$$
$$v_1^2 \le (cx_2 + dx_1)^2 + (cy_2 - dy_1)^2 \\ = c^2 + d^2 + 2cd(x_1 x_2 - y_1 y_2),$$
that is
$$y_1 y_2 - x_1 x_2 \le \frac{c^2 + d^2 - v_1^2}{2cd}. \qquad ②$$
① + ②, we get
$$0 \le \frac{a^2 + b^2 - u^2}{2ab} + \frac{c^2 + d^2 - v_1^2}{2cd},$$
that is
$$\frac{u^2}{ab} + \frac{v_1^2}{cd} \le \frac{a^2 + b^2}{ab} + \frac{c^2 + d^2}{cd}.$$
Similarly
$$\frac{v_1^2}{cd} + \frac{u_1^2}{ab} \le \frac{c^2 + d^2}{cd} + \frac{a^2 + b^2}{ab}.$$
By Cauchy's inequality, we have
$$(u+v)^2 + (u_1+v_1)^2 \\ \le (ab+cd)\left(\frac{u^2}{ab}+\frac{v^2}{cd}\right) + (ab+cd)\left[\frac{u_1^2}{ab}+\frac{v_1^2}{cd}\right] \\ = \frac{u^2}{ab}+\frac{v^2}{cd}+\frac{u_1^2}{ab}+\frac{v_1^2}{cd} \le 2\left[\frac{a^2+b^2}{ab}+\frac{c^2+d^2}{cd}\right].$$

Proof II By Cauchy's inequality, we can show
$$(ay_1 + by_2 + cy_3 + dy_4)^2$$
$$\le (ab + cd) \left[ \frac{(ay_1 + by_2)^2}{ab} + \frac{(cy_3 + dy_4)^2}{cd} \right]$$
$$= \frac{a}{b}y_1^2 + \frac{b}{a}y_2^2 + \frac{c}{d}y_3^2 + \frac{d}{c}y_4^2 + 2(y_1 y_2 + y_3 y_4).$$
Similarly,
$$(ax_4 + bx_3 + cx_2 + dx_1)^2$$
$$\le \frac{a}{b}x_4^2 + \frac{b}{a}x_3^2 + \frac{c}{d}x_2^2 + \frac{d}{c}x_1^2 + 2(x_1 x_2 + x_3 x_4).$$
So we subtract the right-hand side (RHS) from the left-hand side (LHS) in the original inequality and get
$$\begin{align*} \text{LHS} - \text{RHS} &\le \frac{a}{b}y_1^2 + \frac{b}{a}y_2^2 + \frac{c}{d}y_3^2 + \frac{d}{c}y_4^2 + 2y_1 y_2 + 2y_3 y_4 + \\ &\qquad \frac{a}{b}x_4^2 + \frac{b}{a}x_3^2 + \frac{c}{d}x_2^2 + \frac{d}{c}x_1^2 + 2x_1 x_2 + 2x_3 x_4 - \\ &\qquad 2\left(\frac{a}{b} + \frac{b}{a} + \frac{c}{d} + \frac{d}{c}\right) \\ &= -\frac{a}{b}x_1^2 - \frac{b}{a}x_1^2 - \frac{c}{d}x_3^2 - \frac{d}{c}x_4^2 - \frac{a}{b}y_4^2 - \frac{b}{a}y_3^2 - \\ &\qquad \frac{c}{d}y_2^2 - \frac{d}{c}y_1^2 + 2(x_1 x_2 + x_3 x_4 + y_1 y_2 + y_3 y_4) \\ &\le -2x_1 x_2 - 2x_3 x_4 - 2y_1 y_2 - 2y_3 y_4 + \\ &\qquad 2(x_1 x_2 + x_3 x_4 + y_1 y_2 + y_3 y_4) \\ &= 0. \end{align*}$$
The proposition is proved.