(1)n=3
左=(1+2+3)(1+1/2+1/3)=6*(1+1/2+1/3)=6+3+2=11
右=3*3+3-1=11
所以,n=3時不等式成立
(2)假設(shè)n=k(k≥3)時,不等式成立
即(1+2+3+.+k)(1+1/2+.+1/k)≥k²+k-1
當(dāng)n=k+1時,
左=[1+2+3+.+k+1/(k+1)]*[1+1/2+.+1/k+1/(k+1)]
=(1+2+3+.+k)(1+1/2+.+1/k)+(1+2+3+.+k)*(1/k+1)+(k+1)*[1+1/2+.+1/k+1/(k+1)]
≥k²+k-1+k(k+1)/2* (1/k+1)+(k+1)[1+1/2+1/(k+1)]
=k²+k-1+k/2+k+1+(k+1)/2+1
>k²+k-1+k/2+k+1+k/2+1
=k²+3k+1
=(k+1)²+(k+1)-1
所以 n=k+1時,不等式也成立
所以 對大于2的一切正整數(shù)n,不等式都成立