Question

数学 不等式证明 急 求解。

已知x,y,z 是正数。
若 x/(x+2) +y/(y+2) +z/(z+2) =1
求证 x^2/(x+2) +y^2/(y+2) +z^2/(z+2) >=1

Answer 1

还有什么地方不懂的可以再问哈~

Answer 2

x(1/x+2)+y(1/y+2)++z(1/z+2)=1
因为 x，y，z>0
所以
x*x(1/x+2)+y*y(1/y+2)+z*z(1/z+2)>=1

Answer 3

因为x/(x+2) +y/(y+2) +z/(z+2) =1
而1/(x+2) +1/(y+2) +1/(z+2) +(x/(x+2) +y/(y+2) +z/(z+2))/2=3/2
故1/(x+2) +1/(y+2) +1/(z+2) =1
由柯西不等式
1
=x/(x+2) +y/(y+2) +z/(z+2)
=[x/√(x+2) ]*[1/√(x+2)]+[y/√(y+2)]*[1/√(y+2)] +[z/√(z+2)]*[1/√(z+2)]
≤[x^2/(x+2) +y^2/(y+2) +z^2/(z+2)][1/(x+2) +1/(y+2) +1/(z+2)]
=x^2/(x+2) +y^2/(y+2) +z^2/(z+2)
得证

Answer 4

因为原式
所以—