设 $a,b,c,d$ 为正实数，满足 $ab+cd=1$，点 $P_{i}\left(x_{i}, y_{i}\right)(i = 1,2,3,4)$ 是以原点为圆心的单位圆周上的四个点求证：
$ \left(a y_{1}+b y_{2}+c y_{3}+d y_{4}\right)^{2}+\left(a x_{4}+b x_{3}+c x_{2}+d x_{1}\right)^{2} \leqslant 2\left(\frac{a^{2}+b^{2}}{a b}+\frac{c^{2}+d^{2}}{c d}\right)$
$ \left(a y_{1}+b y_{2}+c y_{3}+d y_{4}\right)^{2}+\left(a x_{4}+b x_{3}+c x_{2}+d x_{1}\right)^{2} \leqslant 2\left(\frac{a^{2}+b^{2}}{a b}+\frac{c^{2}+d^{2}}{c d}\right)$
【难度】
【出处】
2003第18届CMO试题
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【答案】
略
【解析】
证法一
令 $u=a y_{1}+b y_{2}, v=c y_{3}+d y_{4}, u_{1}=a x_{4}+b x_{3},v_{1}=c x_{2}+d x_{1}$，则 $u^{2} \leqslant\left(a y_{1}+b y_{2}\right)^{2}+\left(a x_{1}-b x_{2}\right)^{2}=a^{2}+b^{2}+2 a b\left(y_{1} y_{2}-x_{1} x_{2}\right)$ 即 $x_{1} x_{2}-y_{1} y_{2} \leqslant \dfrac{a^{2}+b^{2}-u^{2}}{2 a b}$ ①
$v_{1}^{2} \leqslant\left(c x_{2}+d x_{1}\right)^{2}+\left(c y_{2}-d y_{1}\right)^{2}=c^{2}+d^{2}+2 c d\left(x_{1} x_{2}-y_{1} y_{2}\right)^{2}$ 即 $y_{1} y_{2}-x_{1} x_{2} \leqslant \dfrac{c^{2}+d^{2}-v_{1}^{2}}{2 c d}$ ②
① + ② 得 $0 \leqslant \dfrac{a^{2}+b^{2}-u^{2}}{2 a b}+\dfrac{c^{2}+d^{2}-v_{1}^{2}}{2 c d}$ 即 $\dfrac{u^{2}}{a b}+\dfrac{v_{1}^{2}}{c d} \leqslant \dfrac{a^{2}+b^{2}}{a b}+\dfrac{c^{2}+d^{2}}{c d}$ 同理 $\dfrac{v^{2}}{c d}+\dfrac{u_{1}^{2}}{a b} \leqslant \dfrac{c^{2}+d^{2}}{c d}+\dfrac{a^{2}+b^{2}}{a b}$ 由柯西不等式，有
$(u+v)^{2}+\left(u_{1}+v_{1}\right)^{2} \leqslant (a b+c d)\left(\frac{u^{2}}{a b}+\dfrac{v^{2}}{c d}\right)+(a b+c d)\left(\dfrac{u_{1}^{2}}{a b}+\dfrac{v_{1}^{2}}{c d}\right)=\dfrac{u^{2}}{a b}+\dfrac{v^{2}}{c d}+\dfrac{u_{1}^{2}}{a b}+\dfrac{v_{1}^{2}}{c d} \leqslant 2\left(\dfrac{a^{2}+b^{2}}{a b}+\dfrac{c^{2}+d^{2}}{c d}\right)$
证法二
由柯西不等式可知 $\left(a y_{1}+b y_{2}+c y_{3}+d y_{4}\right)^{2} \leqslant(a b+c d)\left(\dfrac{\left(a y_{1}+b y_{2}\right)^{2}}{a b}+\dfrac{\left(c y_{3}+d y_{4}\right)^{2}}{c d}\right)= \dfrac{a}{b} y_{1}^{2}+\dfrac{b}{a} y_{2}^{2}+\dfrac{c}{d} y_{3}^{2}+\dfrac{d}{c} y_{4}^{2}+2\left(y_{1} y_{2}+y_{3} y_{4}\right)$
同理 $\left(a x_{4}+b x_{3}+c x_{2}+d x_{1}\right)^{2} \leqslant \dfrac{a}{b} x_{4}^{2}+\dfrac{b}{a} x_{3}^{2}+\dfrac{c}{d} x_{2}^{2}+\dfrac{d}{c} x_{1}^{2}+2\left(x_{1} x_{2}+x_{3} x_{4}\right)$
所以，原不等式中
左式－右式 $\leqslant \dfrac{a}{b} y_{1}^{2}+\dfrac{b}{a} y_{2}^{2}+\dfrac{c}{d} y_{3}^{2}+\dfrac{d}{c} y_{4}^{2}+2 y_{1} y_{2}+2 y_{3} y_{4}+ \dfrac{a}{b} x_{4}^{2}+\dfrac{b}{a} x_{3}^{2}+\dfrac{c}{d} x_{2}^{2}+\dfrac{d}{c} x_{1}^{2}+2 x_{1} x_{2}+2 x_{3} x_{4}-2\left(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{c}{d}+\dfrac{d}{c}\right)\\=-\dfrac{a}{b} x_{1}^{2}-\dfrac{b}{a} x_{2}^{2}-\dfrac{c}{d} x_{3}^{2}-\dfrac{d}{c} x_{4}^{2}-\dfrac{a}{b} y_{4}^{2}-\dfrac{b}{a} y_{3}^{2}-\dfrac{c}{d} y_{2}^{2}-\dfrac{d}{c} y_{1}^{2}+2\left(x_{1} x_{2}+x_{3} x_{4}+y_{1} y_{2}+y_{3} y_{4}\right) \\\leqslant-2 x_{1} x_{2}-2 x_{3} x_{4}-2 y_{3} y_{4}-2 y_{1} y_{2}+2\left(x_{1} x_{2}+x_{3} x_{4}+y_{1} y_{2}+y_{3} y_{4}\right )=0$
命题获证．
令 $u=a y_{1}+b y_{2}, v=c y_{3}+d y_{4}, u_{1}=a x_{4}+b x_{3},v_{1}=c x_{2}+d x_{1}$，则 $u^{2} \leqslant\left(a y_{1}+b y_{2}\right)^{2}+\left(a x_{1}-b x_{2}\right)^{2}=a^{2}+b^{2}+2 a b\left(y_{1} y_{2}-x_{1} x_{2}\right)$ 即 $x_{1} x_{2}-y_{1} y_{2} \leqslant \dfrac{a^{2}+b^{2}-u^{2}}{2 a b}$ ①
$v_{1}^{2} \leqslant\left(c x_{2}+d x_{1}\right)^{2}+\left(c y_{2}-d y_{1}\right)^{2}=c^{2}+d^{2}+2 c d\left(x_{1} x_{2}-y_{1} y_{2}\right)^{2}$ 即 $y_{1} y_{2}-x_{1} x_{2} \leqslant \dfrac{c^{2}+d^{2}-v_{1}^{2}}{2 c d}$ ②
① + ② 得 $0 \leqslant \dfrac{a^{2}+b^{2}-u^{2}}{2 a b}+\dfrac{c^{2}+d^{2}-v_{1}^{2}}{2 c d}$ 即 $\dfrac{u^{2}}{a b}+\dfrac{v_{1}^{2}}{c d} \leqslant \dfrac{a^{2}+b^{2}}{a b}+\dfrac{c^{2}+d^{2}}{c d}$ 同理 $\dfrac{v^{2}}{c d}+\dfrac{u_{1}^{2}}{a b} \leqslant \dfrac{c^{2}+d^{2}}{c d}+\dfrac{a^{2}+b^{2}}{a b}$ 由柯西不等式，有
$(u+v)^{2}+\left(u_{1}+v_{1}\right)^{2} \leqslant (a b+c d)\left(\frac{u^{2}}{a b}+\dfrac{v^{2}}{c d}\right)+(a b+c d)\left(\dfrac{u_{1}^{2}}{a b}+\dfrac{v_{1}^{2}}{c d}\right)=\dfrac{u^{2}}{a b}+\dfrac{v^{2}}{c d}+\dfrac{u_{1}^{2}}{a b}+\dfrac{v_{1}^{2}}{c d} \leqslant 2\left(\dfrac{a^{2}+b^{2}}{a b}+\dfrac{c^{2}+d^{2}}{c d}\right)$
证法二
由柯西不等式可知 $\left(a y_{1}+b y_{2}+c y_{3}+d y_{4}\right)^{2} \leqslant(a b+c d)\left(\dfrac{\left(a y_{1}+b y_{2}\right)^{2}}{a b}+\dfrac{\left(c y_{3}+d y_{4}\right)^{2}}{c d}\right)= \dfrac{a}{b} y_{1}^{2}+\dfrac{b}{a} y_{2}^{2}+\dfrac{c}{d} y_{3}^{2}+\dfrac{d}{c} y_{4}^{2}+2\left(y_{1} y_{2}+y_{3} y_{4}\right)$
同理 $\left(a x_{4}+b x_{3}+c x_{2}+d x_{1}\right)^{2} \leqslant \dfrac{a}{b} x_{4}^{2}+\dfrac{b}{a} x_{3}^{2}+\dfrac{c}{d} x_{2}^{2}+\dfrac{d}{c} x_{1}^{2}+2\left(x_{1} x_{2}+x_{3} x_{4}\right)$
所以，原不等式中
左式－右式 $\leqslant \dfrac{a}{b} y_{1}^{2}+\dfrac{b}{a} y_{2}^{2}+\dfrac{c}{d} y_{3}^{2}+\dfrac{d}{c} y_{4}^{2}+2 y_{1} y_{2}+2 y_{3} y_{4}+ \dfrac{a}{b} x_{4}^{2}+\dfrac{b}{a} x_{3}^{2}+\dfrac{c}{d} x_{2}^{2}+\dfrac{d}{c} x_{1}^{2}+2 x_{1} x_{2}+2 x_{3} x_{4}-2\left(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{c}{d}+\dfrac{d}{c}\right)\\=-\dfrac{a}{b} x_{1}^{2}-\dfrac{b}{a} x_{2}^{2}-\dfrac{c}{d} x_{3}^{2}-\dfrac{d}{c} x_{4}^{2}-\dfrac{a}{b} y_{4}^{2}-\dfrac{b}{a} y_{3}^{2}-\dfrac{c}{d} y_{2}^{2}-\dfrac{d}{c} y_{1}^{2}+2\left(x_{1} x_{2}+x_{3} x_{4}+y_{1} y_{2}+y_{3} y_{4}\right) \\\leqslant-2 x_{1} x_{2}-2 x_{3} x_{4}-2 y_{3} y_{4}-2 y_{1} y_{2}+2\left(x_{1} x_{2}+x_{3} x_{4}+y_{1} y_{2}+y_{3} y_{4}\right )=0$
命题获证．
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