若 0<a_ 1 <a_ 2 , 0<b_ 1 <b_ 2…——2008 高考数学第 9 题答案解析

【答案】A

【解析】$a_{1} a_{2}+b_{1} b_{2} \leq\left(\frac{a_{1}+a_{2}}{2}\right)^{2}+\left(\frac{b_{1}+b_{2}}{2}\right)^{2}=\frac{1}{2}$
$a_{1} b_{1}+a_{2} b_{2}-\left(a_{1} b_{2}+a_{2} b_{1}\right)=\left(a_{1}-a_{2}\right) b_{1}+\left(a_{1}-a_{2}\right) b_{2}=\left(a_{2}-a_{1}\right)\left(b_{2}-b_{1}\right) \geq 0$
$a_{1} b_{1}+a_{2} b_{2} \geq\left(a_{1} b_{2}+a_{2} b_{1}\right)$
$1=\left(a_{1}+a_{2}\right)\left(b_{1}+b_{2}\right)=a_{1} b_{1}+a_{2} b_{2}+a_{1} b_{1}+a_{2} b_{1} \leq 2\left(a_{1} b_{2}+a_{2} b_{2}\right)$
$a_{1} b_{1}+a_{2} b_{2} \geq \frac{1}{2}$