若 $0<a_{1}<a_{2,0}, 0<b_{1}<b_{2}$, 且 $a_{1}+a_{2}=b_{1}+b_{2}=1$, 则下列代数式中值最大的是 ( )
方法一 特殊值法.

令 $a_{1}=\frac{1}{4}, a_{2}=\frac{3}{4}, b_{1}=\frac{1}{4}, b_{2}=\frac{3}{4}$,

则 $a_{1} b_{1}+a_{2} b_{2}=\frac{10}{16}=\frac{5}{8}, a_{1} a_{2}+b_{1} b_{2}=\frac{6}{16}=\frac{3}{8}$,

$a_{1} b_{2}+a_{2} b_{1}=\frac{6}{16}=\frac{3}{8}$,
$\because \frac{5}{8}>\frac{1}{2}>\frac{3}{8}, \therefore$ 最大的数应是 $a_{1} b_{1}+a_{2} b_{2}$.

方法二 作差法.

$\because a_{1}+a_{2}=1=b_{1}+b_{2}$ 且 $0<a_{1}<a_{2,0}<b_{1}<b_{2}$,

$\therefore a_{2}=1-a_{1}>a_{1}, \quad b_{2}=1-b_{1}>b_{1}$,

$\therefore 0<a_{1}<\frac{1}{2}, \quad 0<b_{1}<\frac{1}{2}$.

又 $a_{1} b_{1}+a_{2} b_{2}=a_{1} b_{1}+\left(1-a_{1}\right)\left(1-b_{1}\right)=2 a_{1} b_{1}+1-a_{1}-b_{1}$,

$a_{1} a_{2}+b_{1} b_{2}=a_{1}\left(1-a_{1}\right)+b_{1}\left(1-b_{1}\right)=a_{1}+b_{1}-a_{1}^{2}-b_{1}^{2}$,

$a_{1} b_{2}+a_{2} b_{1}=a_{1}\left(1-b_{1}\right)+b_{1}\left(1-a_{1}\right)=a_{1}+b_{1}-2 a_{1} b_{1}$,

$\therefore\left(a_{1} b_{2}+a_{2} b_{1}\right)-\left(a_{1} a_{2}+b_{1} b_{2}\right)=a_{1}^{2}+b_{1}^{2}-2 a_{1} b_{1}$

$=\left(a_{1}-b_{1}\right)^{2} \geqslant 0$,

$\therefore a_{1} b_{2}+a_{2} b_{1} \geqslant a_{1} a_{2}+b_{1} b_{2}$.

$\because\left(a_{1} b_{1}+a_{2} b_{2}\right)-\left(a_{1} b_{2}+a_{2} b_{1}\right)=4 a_{1} b_{1}+1-2 a_{1}-2 b_{1}$

$=1-2 a_{1}+2 b_{1}\left(2 a_{1}-1\right)=\left(2 a_{1}-1\right)\left(2 b_{1}-1\right)$

$=4\left(a_{1}-\frac{1}{2}\right)\left(b_{1}-\frac{1}{2}\right)>0$,

$\therefore a_{1} b_{1}+a_{2} b_{2}>a_{1} b_{2}+a_{2} b_{1}$.

$\because\left(a_{1} b_{1}+a_{2} b_{2}\right)-\frac{1}{2}=2 a_{1} b_{1}+\frac{1}{2}-a_{1}-b_{1}$

$=b_{1}\left(2 a_{1}-1\right)-\frac{1}{2}\left(2 a_{1}-1\right)=\left(2 a_{1}-1\right)\left(b_{1}-\frac{1}{2}\right)$

$=2\left(a_{1}-\frac{1}{2}\right)\left(b_{1}-\frac{1}{2}\right)>0$,

$\therefore a_{1} b_{1}+a_{2} b_{2}>\frac{1}{2}$.

综上可知, 最大的数应为 $a_{1} b_{1}+a_{2} b_{2}$.