证明题;柯西不等式已知x,y,z是正实数,求证:(z^2-x^2)/(x+y)+(x^2-y^2)/(y+z)+(y^2
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/10/05 16:20:04
证明题;柯西不等式
已知x,y,z是正实数,求证:(z^2-x^2)/(x+y)+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)>=0
已知x,y,z是正实数,求证:(z^2-x^2)/(x+y)+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)>=0
证:不妨设x≥y≥z
则x^2≥y^2≥z^2
原式=[(z^2-y^2)+(y^2-x^2)]/(x+y)+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)
=(z^2-y^2)/(x+y)+(y^2-x^2)/(x+y))
+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)
=[(x^2-y^2)/(y+z)+(y^2-x^2)/(x+y)]
+[(y^2-z^2)/(z+x)+(z^2-y^2)/(x+y)
=(x^2-y^2)[1/(y+z)-1/(x+y)]+(y^2-z^2)[1/(z+x)-1/(x+y)]
=(x^2-y^2)·(x-z)/(y+z)(x+y)+(y^2-z^2)·(y-z)/(z+x)((x+y)
=[(x-y)(x-z)]/(y+z)+[(y+z)(y-z)^2]/(z+x)(x+y)
∵ x-y≥0 ,x-z≥0
∴原式≥0
则x^2≥y^2≥z^2
原式=[(z^2-y^2)+(y^2-x^2)]/(x+y)+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)
=(z^2-y^2)/(x+y)+(y^2-x^2)/(x+y))
+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)
=[(x^2-y^2)/(y+z)+(y^2-x^2)/(x+y)]
+[(y^2-z^2)/(z+x)+(z^2-y^2)/(x+y)
=(x^2-y^2)[1/(y+z)-1/(x+y)]+(y^2-z^2)[1/(z+x)-1/(x+y)]
=(x^2-y^2)·(x-z)/(y+z)(x+y)+(y^2-z^2)·(y-z)/(z+x)((x+y)
=[(x-y)(x-z)]/(y+z)+[(y+z)(y-z)^2]/(z+x)(x+y)
∵ x-y≥0 ,x-z≥0
∴原式≥0
已知 x,y,z都是正实数,且 x+y+z=xyz 证明 (y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1
不等式证明 急 已知x,y,z 是正数.若 x/(x+2) +y/(y+2) +z/(z+2) =1求证 x^2/(x+
x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)
一道高中不等式证明题已知正数x,y,z满足x+y+z=1求证:x^2/(y+2z)+y^2/(z+2x)+z^2/(x+
x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)
若x,y,z是正实数,且x+y+z=xyz,证明:(y+z/x)+(z+x/y)+(x+y/z)≥2倍的(1/x)+(1
已知三个实数x,y,z满足条件(z-x)^2-4(x-y)(y-z)=0,求证:x,y,z成等差数列
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
已知x,y,z 大于0,x+y+z=2,求证 xz/y(y+z)+zy/x(x+y)+yx/z(z+x)大于等于2/3
(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z
已知(x+y+z)^2=x^2+y^2+z^2,证明x(y+z)+y(z+x)+z(x+y)=0
已知x,y,z属于R+(正实数),且xyz(x+y+z)=4+2*根号下3,则(x+y)(y+z)的最小值是?