证明
由n^3/m+m^3/n-m^2-n^2
=n^3/m-m^2+m^3/n-n^2
=(n^3-m^3)/m+(m^3-n^3)/n
=(m^3-n^3)(1/n-1/m)
=(m^3-n^3)(m-n)/mn
=(m-n)(m^2+mn+n^2)(m-n)/mn
=(m-n)^2(m^2+mn+n^2)/mn
=(m-n)^2[(m-1/2n)^2+3n^2/4]/mm
由m,n是正整数
知(m-n)^2≥0,[(m-1/2n)^2+3n^2/4]>0,mm>0
即(m-n)^2[(m-1/2n)^2+3n^2/4]/mm≥0
故n^3/m+m^3/n≥m^2+n^2