A解析:一:特值法,不妨令a1=0.2,a2=0.8,b1=0.3,b2=0.7,
逐一代入可得.
解析二:由A项减B项有:a1b1+a2b2-(a1a2+b1b2)=(b1-a2)(a1-b2),
又由0<a1<a2,a1+a2=1知0<a1<,≤a2<1.
同理有0<b1<,≤b2<1.
∴(b1-a2)(a1-b2)>0.
∴a1b1+a2b2>a1a2+b1b2.①
又知a1a2+b1b2<+=,②
∴a1b1+a2b2>.③
又知C项a1b2+a2b1=a1(1-b1)+a2(1-b2)=a1+a2-(a1b1+a2b2)=1-(a1b1+a2b2)<,
∴A项代数式值最大.