定理　设 0 a1≤a2 ≤…≤an,0 b1≤b2 ≤…≤bn,k1,… ,kn 是 1 ,2 ,…n的任一排列 ,则∑nj=11aj+bn + 1-j≤∑nj=11aj+bkj≤∑nj=11aj+bj(a1=an,b1=bn)①(简记为反序和≤乱序和≤同序和 ) .证明 :记①式的中间和为Sn,仅证①式右边不等式 ,即证明当kj=j(j =1 ,… ,n)时 ,S Theorem is set to 0 a1 ≤ a2 ≤ ... ≤ an, 0 b1 ≤ b2 ≤ ... ≤ bn, k1, ... ,kn is any arrangement of 1, 2, ... n, then ∑nj = 11aj + bn + 1-j ≤ ∑ Nj=11aj+bkj≤∑nj=11aj+bj(a1=an,b1=bn)1 (remember to be reversed and ≤in disorder and ≤ordinal sum). Prove that the middle sum of note 1 is Sn Only the right inequality of type 1 is proved, that is, when kj=j(j =1,...,n), S