已知:x⼀(y+z )+y⼀(x+z )+z⼀(x+y )=1 求x^2⼀(y+z) + y^2⼀(x+z) +z^2⼀(x+y)的值

x/(y+z )+y/(x+z )+z/(x+y )=1
等式分别乘以x，y，z
得到
x^2/(y+z)+ xy/(x+z)+xz/(x+y)=x
xy/（y+z）+y^2/(x+z)+yz/(x+y)=y
xz/(y+z)+yz/(x+z)+z^2/(x+y)=z
三个式子相加
得到
x^2/(y+z) + y^2/(x+z) +z^2/(x+y)+ xy/(x+z)+xz/(x+y)+xy/（y+z)+yz/(x+y)xz/(y+z)+yz/(x+z)=x+y+z
令x^2/(y+z) + y^2/(x+z) +z^2/(x+y)=a
那么上面的式子变成
a+（xy+yz）/(x+z)+(xz+yz)/(x+y)+(yz+zx)/(y+z)=x+y+z
提取公因式后可以约分
a+y+z+x=x+y+z
所以a=0
即x^2/(y+z) + y^2/(x+z) +z^2/(x+y)=0

这个条件有问题的
利用柯西不等式
[x/(y+z)+y/(x+z)+z/(x+y)]*[x(y+z)+y(x+z)+z(x+y)>=(x+y+z)^2
化简下即
2xy+2yz+2xz>=x^2+y^2+z^2+2xy+2yz+2xz
x^2+y^2+z^2<=0
x=y=z=0
但是这样，原式分母为0，矛盾

由x/(y+z )+y/(x+z )+z/(x+y )=1 知

x/(y+z)=1-y/(x+z)-z/(x+y)
y/(x+z)=1-x/(y+z)-z/(x+y)
z/(x+y)=1-x/(y+z)-y/(x+z)
所以
x^2/(y+z)=x-xy/(x+z)-xz/(x+y)
y^2/(x+z)=y-yx/(y+z)-yz/(x+y)
z^2/(x+y)=z-zx/(y+z)-zy/(x+z)

以上三个相加
x^2/(y+z) + y^2/(x+z) +z^2/(x+y)
=x-xy/(x+z)-xz/(x+y)+y-yx/(y+z)-yz/(x+y)+z-zx/(y+z)-zy/(x+z)
=x+y+z-x(y+z)/(y+z)-y(x+z)/(x+z)-z(x+y)/(x+y)
=x+y+z-x-y-z
=0

令x+y+z=a
原方程化为
x/(a-x)+y/(a-y)+z/(a-z)=1
a/(a-x)-1+a/(a-y)-1+a/(a-z)-1=1
1/(a-x)+1/(a-y)+1/(a-z)=4/a
而所求的
x^2/(y+z) + y^2/(x+z) +z^2/(x+y)
=x^2/(a-x)+y^2/(a-y)+z^2/(a-z)
=a^2/(a-x)-a-x+a^2/(a-y)-a-y+a^2/(a-z)-a-z
=a^2*(1/(a-x)+1/(a-y)+1/(a-z))-4a
=a^2*4/a-4a
=0