高中不等式证明a,b,c均大于等
令a=x^3 b=y^3 c=z^3
x^3y^3z^3=1
xyz=1
x^2-2xy+y^2>=0
(x+y)(x^2-xy+y^2)>=xy(x+y)
x^3+y^3>=xy(x+y)
x^3+y^3+xyz>=xy(x+y+z)
1/(x^3+y^3+xyz)<=1/[xy(x+y+z)]
同理，1/(x^3+y^3+xyz)<=1/[yz(x+y+z)]
1/(x^3+y^3+xyz)<=1/[xz(x+y+z)]
原式=xyz/(x^3+y^3+xyz)+xyz/(y^3+z^3+xyz)+xyz/(x^3+z^3+xyz)
<=xyz/[xy(x+y+z)]+xyz/[yz(...全部
令a=x^3 b=y^3 c=z^3
x^3y^3z^3=1
xyz=1
x^2-2xy+y^2>=0
(x+y)(x^2-xy+y^2)>=xy(x+y)
x^3+y^3>=xy(x+y)
x^3+y^3+xyz>=xy(x+y+z)
1/(x^3+y^3+xyz)<=1/[xy(x+y+z)]
同理，1/(x^3+y^3+xyz)<=1/[yz(x+y+z)]
1/(x^3+y^3+xyz)<=1/[xz(x+y+z)]
原式=xyz/(x^3+y^3+xyz)+xyz/(y^3+z^3+xyz)+xyz/(x^3+z^3+xyz)
<=xyz/[xy(x+y+z)]+xyz/[yz(x+y+z)]+xyz/[xz(x+y+z)]
=z/(x+y+z)+x/(x+y+z)+y/(x+y+z)
=1
所以原式<=1
。
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