早教吧作业答案频道 -->数学-->
求证当x+y+z=1√(x∧2+y∧2+z∧2+xy+yz+xz)≥√6/3不可以用柯西不等式用基本不等式证明
题目详情
求证 当x+y+z=1 √(x∧2+y∧2+z∧2+xy+yz+xz) ≥√6/3 不可以用柯西不等式 用基本不等式证明
▼优质解答
答案和解析
证明:x+y+z=1
(x+y+z)²=1=x²+y²+z²+2xy+2xz+2yz
x²+xy+y²+y²+yz+z²+ z²+zx+x²
=x²+y²+z²+2xy+2xz+2yz+(x²+y²+z²-xy-xz-yz)
=1+(x²+y²+z²-xy-xz-yz)
=1+½(2x²+2y²+2z²-2xy-2xz-2yz)
=1+½[(x-y)²+(y-z)²+(z-x)²]≥3/2
当且仅当x=y=z=1/3等号成立
∴√(x²+y²+z²+xy+yz+xz) ≥√6/3
得证
(x+y+z)²=1=x²+y²+z²+2xy+2xz+2yz
x²+xy+y²+y²+yz+z²+ z²+zx+x²
=x²+y²+z²+2xy+2xz+2yz+(x²+y²+z²-xy-xz-yz)
=1+(x²+y²+z²-xy-xz-yz)
=1+½(2x²+2y²+2z²-2xy-2xz-2yz)
=1+½[(x-y)²+(y-z)²+(z-x)²]≥3/2
当且仅当x=y=z=1/3等号成立
∴√(x²+y²+z²+xy+yz+xz) ≥√6/3
得证
看了求证当x+y+z=1√(x∧2...的网友还看了以下:
化简[x^3(y^2-z^2)+y^3(z^2-x^2)+z^3(x^2-y^2)]/[x^3(y 2020-06-03 …
xyz=1,x+y+z=2,x^2+y^2+z^2=3,求x,y,z我解:xy=1/z,x+y=2- 2020-10-31 …
x,y,z>0x^2/(1+x^2)+y^2/(1+y^2)+z^2/(1+z^2)=2求证x/(1 2020-10-31 …
1.x+y+z≠0且x/(y+z)=y/(x+y)=z/x+y,求x/(x+y+z)2.x+y+z= 2020-10-31 …
用matlab解这样一个方程组怎么解不出来啊[x,y,z]=solve('x^2+y^2+z^2=r 2020-10-31 …
设正数x,y,z,满足不等式:x^2+y^2-z^2/2xy+y^2+z^2-x^2/2yz+z^2 2020-11-01 …
(a+b+c)/3大于等于3*√abc设a=x^3,b=y^3,c=z^3x,y,z是非负数时x^3 2020-11-01 …
由(x^2+y^2+z^2)*(x+y+z)=x^3+y^3+z^3+(x+y)z^2+(y+z)x 2020-11-01 …
设实数x>0,y>0,z>0,a>0,b>0,且x,y,z满足条件x^2+y^2-xy=a^2;x^ 2020-11-01 …
下列各式中,与(x-y+z)(x+y-z)相等的是()A.x^2-y^2-z^2B.-(x+y+z) 2020-11-01 …