(1) When n=3, the proposition holds.

(2) Suppose that when n=k, the proposition holds, that is, (1+2+...+k) (1+1+...1/k) = k 2+k-/kloc-.

When n=k+ 1, [1+2+...+k+k+1] [1+1/2+...1/k+/kloc.

=( 1+2+...+k)( 1+ 1/2+.... 1/k)+( 1+2+...+k)* 1/(k+ 1)+( 1+ 1/2+.... 1/k)*(k+ 1)+ 1

& gt=(k^2+k- 1)+k/2+( 1+ 1/2)*(k+ 1)+ 1

=k^2+3k+3/2

& gt=k^2+3k+ 1

=k^2+2k+ 1+k+ 1- 1

=(k+ 1)^2+(k+ 1)- 1

So when n=k+ 1, the inequality holds.

(1)(2) Know the original inequality.