若xyz属于r 求证 x2+y2+z2大于等于xy+yz+zx

若xyz属于r 求证 x2+y2+z2大于等于xy+yz+zx 已知x,y,z∈R+.求证(1+x2)(1+y2)(1+z2)≥8xyz 已知xyz满足z+y+z=xyz 求证：x(1-y2)(1-z2)+y(1-x2)(1-z2)+z(1-x2)(1-y2)=4xyz 设x,y,z∈R+,求证 2z2-x2-y2/(x+y)+2x2-y2-z2/(y+z)≥x2+z2-2y2/(x+z) 设x,y,z∈R+,求证 2z2-x2-y2/(x+y)+2x2-y2-z2/(y+z)≥x2+z2-2y2/(x+z) 已知x,y,z互不相等,且xyz不等于0,x2＋yz＝z2,y2＋zx＝x2,求证：z2＋xy＝y2 高中反三角函数问题arccosx+arccosy+arccosz=∏,求证x2+y2+z2+2xyz=1 已知x+y+z=xyz,证明：x(1-y2)(1-z2)+y(1-x2)(1-z2)+z(1-x2)(1-y2)=4xyz 已知（x2+y2+z2）（a2+b2+c2）=（ax+by+cz）2,xyz≠0,求证a/x=b/y=c/z X2+Y2+Z2 已知X,Y,Z∈R,且X+Y+Z=1,求证X2+Y2+Z2≥1/3 设x,y,z属于R,且（2x+y-z)/(根号下（X2+2y2+z2))的最大值 求证x2+y2+z2>=(x+y+z)平方/3 x,y,z为正实数 求证 x2/(y2+z2+yz)+y2/(z2+x2+zx)+z2/(x2+y2+xy)>=1 因式分解X2(Y+Z)+Y2(Z+X)+Z2(X+Y)-(X3+Y3+Z3)-2XYZ 不等式问题:正实数x,y,z满足xyz≥1,证明(x5-x2)/(x5+y2+z2)+(y5-y2)/(y5+z2+x2)+(z5-z2)/(z5+x2+y2)≥0字母右边的数字是指数,应该是用柯西不等式的 求方程xyz + x2 + y2 + z2 = 2 确定的函数z = z( x,y)在点(1,0,-1)处的全微分dz,是xyz + （x2 + y2 + z2）^(1/2) =2^(1/2) 已知x+y+z=3,x2+y2+z2=19,x3+y3+z3=30则xyz=?