若a1≤a2≤…≤an,b1≤b2≤…≤bn,则有(a1b1+a2b2+…+anbn)≥ (a1+a2+…+an)(b1+b

a1a2≤…≤an,b1b2≤…≤bn,则有(a1b1+a2b2+…+anbn)≥ (a1+a2+…+an)

(b1+b2+…+bn).

答案

证明:

S=a1b1+a2b2+…+anbn是同序和,则

Sa1b2+a2b3+…+anb1,

Sa1b3+a2b4+…+anb2,

……

Sa1b1+a2b2+…+anbn.

将上面几个式子相加,并按列求和得

nSa1(b1+b2+…+bn)+a2(b1+b2+…+bn)+…+an(b1+b2+…+bn)=(a1+a2+…+an)(b1+b2+…+bn).

所以

S(a1+a2+…+an)(b1+b2+…+bn),

a1b1+a2b2+…+anbn(a1+a2+…+an)(b1+b2+…+bn).

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