- 不等式难题求助大家设x,y,z为非负实数，且x^2+y^2+z^
- 设x,y,z为非负实数，且x^2+y^2+z^2=1.求证
见图解.
并要说明不等式成立的条件.
- ∵x^2+y^2+z^2=1，∴首先齐次化为
∑(x^2+y^2+z^2)/(x^2+y^2+z^2-yz)≤4∑x*∑yz/∏(y+z).
借助于Maple,去分母展开整理
eand(4*(x+y+z)*(y*z+z*x+x*y)*(x^2+y^2+z^2-y*z)*(x^2+y^2+z^2-z*x)*(x^2+y^2+z^2-x*y)-(x^2+y^2+z^2)*(3*(x^2+y^2+z^2)^2-2*(y*z+z*x+x*y)*(x^2+y^2+z^2)+x*y*z*(x+y+z))*(y+z)*(z+x)*(x+y));
∑yz(y^7+z^7)-∑(yz)^2*(y^5+z^5)+∑(yz)^3*(y^3+z^3)-∑(yz)^4*(y+z)
+xyz[2∑x^6-5∑(y+z)x^5+12∑(y^2+z^2)x^4-12∑(yz)^3]
+(xyz)^2*[2∑x^3-6∑(y+z)x^2+18xyz]≥0
上式化简为
∑yz[2x^4*(y+z)-3x^3*yz+x^2*(y+z)+x(y^4-3zy^3+5y^2*z^2-3yz^3+z^4)
+y^5+zy^4+2y^3*z^2+2y^2*z^2+yz^4+z^5]*(y-z)^2≥0
＜＝＝＞
∑yz[x^2*y(x-z)^2+x^2*z(x-y)^2+x(y^2-yz+z^2)*(y-z)^2+x^4*(y+z)+x^3*yz
+xy^2*z^2+y^5+zy^4+2y^3*z^2+2y^2*z^2+yz^4+z^5]*(y-z)^2≥0
当x=y=z时,或x=0,y=z=√(1/2)时取等号。