求双曲题设x,y,z为正实数,满足:yz+zx+xy=1,求证:题设x,y,z为正实数,满足:yz+zx+xy=1,求证:27(y+z)(z+x)(x+y)/4≥[√(y+z)+√(z+x)+√(x+y)]^2≥6√3.

Question

求双曲题 设x,y,z为正实数,满足:yz+zx+xy=1,求证:
题 设x,y,z为正实数,满足:yz+zx+xy=1,求证:
27(y+z)(z+x)(x+y)/4≥[√(y+z)+√(z+x)+√(x+y)]^2≥6√3.

Answer 1

设x,y,z为正实数,且满足:yz+zx+xy=1,求证:
27(y+z)(z+x)(x+y)/4≥[√(y+z)+√(z+x)+√(x+y)]^2≥6√3.
证明 根据均值不等式得
[√(y+z)+√(z+x)+√(x+y)]^2≤6(x+y+z) ,
所以欲证不等式链左边,只需证
9(y+z)(z+x)(x+y)≥8(x+y+z) (1)
(1) 9(y+z)(z+x)(x+y)≥8(x+y+z)*( yz+zx+xy)
x*(y-z)^2+y*(z-x)^2+z*(x-y)^2≥0,
(1) 显然成立.故不等式链左边成立
据三元均值不等式
[√(y+z)+√(z+x)+√(x+y)]^2=
2(x+y+z)+2√[(z+x)(x+y)]+2√[(x+y)(y+z)]+2√[(y+z)(z+x)]
≥2(x+y+z)+6[(y+z)(z+x)(x+y)]^(1/3)
≥2(x+y+z)+6[8(x+y+z)/9]^(1/3)
注意由yz+zx+xy=1可导出:x+y+z≥√3,所以
2(x+y+z)+6[8(x+y+z)/9]^(1/3)≥2√3+6*[(8√3)/9]^(1/3)
=2√3+6*(2/√3)=6√3.故不等式链右边成立.证毕.