- 请教一道不等式题目x+y+z=1，x,y,z∈R+证明:9xyz
- x+y+z=1，x,y,z∈R+
证明:9xyz+1≥4ab+4bc+4ca
- x+y+z=1，x,y,z∈R+
证明:9xyz+1≥4ab+4bc+4ca
符号有误,应该为4(yz+zx+xy)
9xyz+1≥4(yz+zx+xy)
因为x+y+z=1,对所证不等式齐次化,等价于
9xyz+(x+y+z)^3≥4(yz+zx+xy)(x+y+z)
<===>
x^3+y^3+z^3-(y+z)x^2-(y^2+z^2)x-yz(y+z)+3xyz≥0
不失一般性,设x=min(x,y,z),则上式分解为
x(x-y)(x-z)+(y+z-x)(y-z)^2≥0
显然成立.
其实上述代数不等式等价于几何不等式:
s^2≥16Rr+5r^2.