數(shù)歸法
1、當(dāng)n=3時(shí) 左邊=11 右邊=11,顯然成立
2、假設(shè) n=k時(shí)成立,有
(1+2+3+……+k）(1+1/2+1/3+……+1/k）=k方+k-1
那么
左邊=[1+2+3+……+k+(k+1)][1+1/2+1/3+……+1/k +1/(k+1)]
=[(k+1)(k+2)/2][1+1/2+...+1/k +1/(k+1)]
=[(k+2)/k][k(k+1)/2](1+1/2+...+1/k) + (k+2)/2
>= [(k+2)/k](k^2+k-1) + (k+2)/2
= [(k+1)^2+(k+1)-1] +(k^2+2k-4)/2k
>= (k+1)^2+(k+1)-1 =右邊,不等式成立
所以 當(dāng)n屬于正整數(shù)時(shí) n=k+1時(shí)成立