不等式 2道若X1 X2 X3属于R y1 y2 y3属于R+证明 (x1)^2/y1+(x2)^2/y2+(x3)^2/y3>=(x1+x2+x3)^2/(y1+y2+y3)当且仅当x1/y2=x2/y2=x3/y3 成立已知x .y.z 属于R+ 且 xyz=1 求x^2/(y+z)+y2/(z+x)+z^2/(x+y)最小值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/29 22:23:30
![不等式 2道若X1 X2 X3属于R y1 y2 y3属于R+证明 (x1)^2/y1+(x2)^2/y2+(x3)^2/y3>=(x1+x2+x3)^2/(y1+y2+y3)当且仅当x1/y2=x2/y2=x3/y3 成立已知x .y.z 属于R+ 且 xyz=1 求x^2/(y+z)+y2/(z+x)+z^2/(x+y)最小值](/uploads/image/z/13317893-53-3.jpg?t=%E4%B8%8D%E7%AD%89%E5%BC%8F+2%E9%81%93%E8%8B%A5X1+X2+X3%E5%B1%9E%E4%BA%8ER+y1+y2+y3%E5%B1%9E%E4%BA%8ER%2B%E8%AF%81%E6%98%8E+%28x1%29%5E2%2Fy1%2B%28x2%29%5E2%2Fy2%2B%28x3%29%5E2%2Fy3%3E%3D%28x1%2Bx2%2Bx3%29%5E2%2F%28y1%2By2%2By3%29%E5%BD%93%E4%B8%94%E4%BB%85%E5%BD%93x1%2Fy2%3Dx2%2Fy2%3Dx3%2Fy3+%E6%88%90%E7%AB%8B%E5%B7%B2%E7%9F%A5x+.y.z+%E5%B1%9E%E4%BA%8ER%2B+%E4%B8%94+xyz%3D1+%E6%B1%82x%5E2%2F%28y%2Bz%29%2By2%2F%28z%2Bx%29%2Bz%5E2%2F%28x%2By%29%E6%9C%80%E5%B0%8F%E5%80%BC)
不等式 2道若X1 X2 X3属于R y1 y2 y3属于R+证明 (x1)^2/y1+(x2)^2/y2+(x3)^2/y3>=(x1+x2+x3)^2/(y1+y2+y3)当且仅当x1/y2=x2/y2=x3/y3 成立已知x .y.z 属于R+ 且 xyz=1 求x^2/(y+z)+y2/(z+x)+z^2/(x+y)最小值
不等式 2道
若X1 X2 X3属于R y1 y2 y3属于R+
证明 (x1)^2/y1+(x2)^2/y2+(x3)^2/y3>=(x1+x2+x3)^2/(y1+y2+y3)
当且仅当x1/y2=x2/y2=x3/y3 成立
已知x .y.z 属于R+ 且 xyz=1 求x^2/(y+z)+y2/(z+x)+z^2/(x+y)最小值
不等式 2道若X1 X2 X3属于R y1 y2 y3属于R+证明 (x1)^2/y1+(x2)^2/y2+(x3)^2/y3>=(x1+x2+x3)^2/(y1+y2+y3)当且仅当x1/y2=x2/y2=x3/y3 成立已知x .y.z 属于R+ 且 xyz=1 求x^2/(y+z)+y2/(z+x)+z^2/(x+y)最小值
1.
Cauchy不等式
(x1^2/y1+x2^2/y2+x3^2/y3)(y1+y2+y3)
≥(x1+x2+x3)^2
得证
2.
还是Cauchy
(x^2/(y+z)+y^2/(x+z)+z^2/(x+y))(2x+2y+2z)
≥
(x+y+z)^2
所以原式
≥(x+y+z)/2
≥3(xyz)^(1/3)/2
=3/2
最小值3/2
1.由权方和不等式知为显然。
2.由权方和不等式可得:
x^2/(y+z)+y2/(z+x)+z^2/(x+y)>=(x+y+z)^2/(y+z+z+x+x+y)=(x+y+z)/2>=3三次根号(xyz)/2=3/2
最小值为3/2