(n!)^2=(n*(n-1)*...*1)^2 =(n*1)*((n-1)*2)*((n-2)*3)*...*(1*n)往证(n+1-i)*i>=n即可, (1<=i<=n)即i^2-(n+1)*i+n<=0即(i-1)(i-n)<=0这显然成立
