分类：教育/科学-升学入学-高考
问题：不等式证明设x,y,z&gt;0,且x^2+y^2+z^2=1. 
设x,y,z>0,且x^2+y^2+z^2=1.求证
1/(1-yz)+1/(1-zx)+1/(1-xy)=<
4(x+y+z)*(yz+zx+xy)/[(y+z)(z+x)(x+y)]
回答：参考文献： 据题设条件:x^2+y^2+z^2=1，首先对所证不等式齐次化为 ∑(x^2+y^2+z^2)/(x^2+y^2+z^2-yz)≤4∑x*∑yz/∏(y+z). 上式去分母,展开整理为 ∑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[yx^2(x-z)^2+zx^2(x-y)^2+x(y^2-yz+z^2)(y-z)^2+(y+z)x^4+yzx^3 +x(yz)^2+y^5+zy^4+2y^3*z^2+2y^2*z^3+yz^4+z^5](y-z)^2≥0 易验证当 x=y=z,或 x=0,y=z=√(1/2)时,等号成立。