a=x^3 b=y^3 c=z^3x^3+y^3+z^3-3xyz =[( x+y)^3-3x^2y-3xy^2]+z^3-3xyz=[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z) =(x+y+z)(x^2+y^2+2xy-xz-yz+z^2)-3xy(x+y+z) =(x+y+z)(x^2+y^2+z...怎么證明任意N(>3)個(gè)數(shù)的算術(shù)平均數(shù)恒大于等于它們的幾何平均數(shù)？那就麻煩了，基本上得用冪平均不等式或者琴生不等式其實(shí)構(gòu)造函數(shù)也行，也難再就是數(shù)學(xué)歸納法，反向歸納好一些