xyz都是正数,且x²/(1+x²)+y²/(1+y²)+z²/(1+
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/04/28 15:36:56
xyz都是正数,且x²/(1+x²)+y²/(1+y²)+z²/(1+z²)=2,求证 x/(1+x²)+y/(1+y²)+z/(1+z²)<=根号2
x²/(1+x²)+y²/(1+y²)+z²/(1+z²)=2
则[1-1/(1+x²)]+[1-1/(1+y²)]+[1-1/(1+z²)]=2
所以1/(1+x²)+1/(1+y²)+1/(1+z²)=1
由柯西不等式
2=[x²/(1+x²)+y²/(1+y²)+z²/(1+z²)][1/(1+x²)+1/(1+y²)+1/(1+z²)]
≥[x/(1+x²)+y/(1+y²)+z/(1+z²)]²
所以x/(1+x²)+y/(1+y²)+z/(1+z²)≤√2
则[1-1/(1+x²)]+[1-1/(1+y²)]+[1-1/(1+z²)]=2
所以1/(1+x²)+1/(1+y²)+1/(1+z²)=1
由柯西不等式
2=[x²/(1+x²)+y²/(1+y²)+z²/(1+z²)][1/(1+x²)+1/(1+y²)+1/(1+z²)]
≥[x/(1+x²)+y/(1+y²)+z/(1+z²)]²
所以x/(1+x²)+y/(1+y²)+z/(1+z²)≤√2
1、x³+x²y-x²z-xyz
已知x,y,z都是正数,且xyz=1,求证:x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥3/2
已知x,y,z都是正数,且xyz=1,求证:xy(x+y)+yz(y+z)+zx(z+x)》6
数学题已知xyz≠0,且x+y+z=1,x²+y²+z²=1,求1/x+1/y+1/z的值
已知正数xyz,满足x+y+z=xyz 已知正数x,y,z满足x+y+z=xyz,且不等式1/x+y+1/y+z+1/z
xyz≠0,且x+y+z=0,求证根号(1/x²+1/y²+1/z²)=(1/x+1/y+
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
已知x ,y ,z都是正数且满足xyz(x+y+z)=1试求(x+y)(y+z)取得最小值时x,y,z的值各是多少?
知x,y,z都是正数,且x+y+z=xyz,求1/根号xy+1/根号yz+2/根号xz的最大值
x+y+z=1,x,y,z都是正数,求xy+yz+xz-3xyz的最大值和最小值
己知x,y,z都是非零有理数,且满足|x|/x+|y|/y+z/|z|=1,请你求xyz/|xyz|的值.求因为所以?
己知x,y, z都是非零有理数,且满足|x|/x+|y|/y+z/|z|=1,请你求xyz/|x