已知x,y,z为任意实数,满足xy+yz+zx=1,求证xyz(x+y+z)≤1⼀3

2025-05-01 19:48:32
推荐回答(2个)
回答1:

(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)=1
x^2y^2+y^2z^2+z^2x^2=1/3(x^2y^2+y^2z^2+z^2x^2)(1+1+1)≥1/3(xy+yz+zx)^2=1/3(柯西不等式)
所以xyz(x+y+z)≤1/3

回答2:

看图