分类：教育/科学-理工学科-数学
问题：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.
回答：∵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。