China Mathematical Olympiad

Proof I Set $u=a{y}_1+b{y}_2$, $v=c{y}_3+d{y}_4$, $u_1=a{x}_4+b{x}_3$ and $v_1=c{x}_2+d{x}_1$. Then $$\begin{gathered} u^2\le (a{y}_1+b{y}_2{)}^2+(a{x}_1-b{x}_2{)}^2 \\ =a^2+b^2+2ab(y_1{y}_2-x_1{x}_2), \end{gathered}$$ that is $$x_1{x}_2-y_1{y}_2\le \frac{a^2+b^2-u^2}{2ab}.①$$ $$\begin{gathered} v_1^2\le (c{x}_2+d{x}_1{)}^2+(c{y}_2-d{y}_1{)}^2 \\ =c^2+d^2+2cd(x_1{x}_2-y_1{y}_2), \end{gathered}$$ that is $$y_1{y}_2-x_1{x}_2\le \frac{c^2+d^2-v_1^2}{2cd}.②$$ 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 $$\begin{gathered} (u+v{)}^2+(u_1+v_1{)}^2 \\ \le (ab+cd)\left(\frac{u^2}{ab}+\frac{v^2}{cd}\right)+(ab+cd)\left[\frac{u_1^2}{ab}+\frac{v_1^2}{cd}\right] \\ =\frac{u^2}{ab}+\frac{v^2}{cd}+\frac{u_1^2}{ab}+\frac{v_1^2}{cd}\le 2\left[\frac{a^2+b^2}{ab}+\frac{c^2+d^2}{cd}\right]. \end{gathered}$$

Proof II By Cauchy's inequality, we can show $$(a{y}_1+b{y}_2+c{y}_3+d{y}_4{)}^2$$ $$\le (ab+cd)\left[\frac{(a{y}_1+b{y}_2{)}^2}{ab}+\frac{(c{y}_3+d{y}_4{)}^2}{cd}\right]$$ $$=\frac{a}{b}{y}_1^2+\frac{b}{a}{y}_2^2+\frac{c}{d}{y}_3^2+\frac{d}{c}{y}_4^2+2(y_1{y}_2+y_3{y}_4).$$ Similarly, $$(a{x}_4+b{x}_3+c{x}_2+d{x}_1{)}^2$$ $$\le \frac{a}{b}{x}_4^2+\frac{b}{a}{x}_3^2+\frac{c}{d}{x}_2^2+\frac{d}{c}{x}_1^2+2(x_1{x}_2+x_3{x}_4).$$ So we subtract the right-hand side (RHS) from the left-hand side (LHS) in the original inequality and get $$\begin{array}{rl}\text{LHS}-\text{RHS}& \le \frac{a}{b}{y}_1^2+\frac{b}{a}{y}_2^2+\frac{c}{d}{y}_3^2+\frac{d}{c}{y}_4^2+2y_1{y}_2+2y_3{y}_4+\\ & \frac{a}{b}{x}_4^2+\frac{b}{a}{x}_3^2+\frac{c}{d}{x}_2^2+\frac{d}{c}{x}_1^2+2x_1{x}_2+2x_3{x}_4-\\ & 2\left(\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\right)\\ & =-\frac{a}{b}{x}_1^2-\frac{b}{a}{x}_1^2-\frac{c}{d}{x}_3^2-\frac{d}{c}{x}_4^2-\frac{a}{b}{y}_4^2-\frac{b}{a}{y}_3^2-\\ & \frac{c}{d}{y}_2^2-\frac{d}{c}{y}_1^2+2(x_1{x}_2+x_3{x}_4+y_1{y}_2+y_3{y}_4)\\ & \le -2x_1{x}_2-2x_3{x}_4-2y_1{y}_2-2y_3{y}_4+\\ & 2(x_1{x}_2+x_3{x}_4+y_1{y}_2+y_3{y}_4)\\ & =0.\end{array}$$ The proposition is proved.