Question

用比较发证明,若a,b,c都是正实数,则a^3+b^3+c^3>=3abc

Answer 1

分解因式

a^3+b^3+c^3-3abc

= (a+b+c)*(a^2+b^2+c^2-ab-bc-ac)

= (a+b+c)*[(a-b)^2+(a-c)^2+(b-c)^2]/2

a+b+c>=0

[(a-b)^2+(a-c)^2+(b-c)^2]>=0

故a^3+b^3+c^3-3abc>=0

即a^3+b^3+c^3>=3abc