不等式的证明x,y,z为正的实数求证（xy+yz+zx)[1/(x+y)^2+1/(y+z)^2+1/(x+z)^2]>=9/4

Question

不等式的证明
x,y,z为正的实数
求证
（xy+yz+zx)[1/(x+y)^2+1/(y+z)^2+1/(x+z)^2]>=9/4

Answer 1

s=(3/2)^2-4(xy/4+yz/4+zx/4)[1/(x+y)^2+1/(y+z)^2+1/(x+z)^2]
这就让我们想到用构造函数法来证,虽麻烦了点,但思路确是很清晰的
f(t)=[1/(x+y)^2+1/(y+z)^2+1/(x+z)^2]*t^2-3/2*t+(xy/4+yz/4+zx/4)
=[1/(x+y)^2*t^2-1/2*t+xy/4]
+[1/(y+z)^2*t^2-1/2*t+yz/4]
+[1/(x+z)^2*t^2-1/2*t+zx/4]
因为(x+y)^2>=4xy
(1/2)^2-4*(xy/4)*1/(x+y)^2=1/4-xy/(x+y)^2