设(y+z)⼀a^2=(z+x)⼀b^2=(x+y)⼀c^2,且xy+yz+zx=0,xyz≠0,求证:(a+b+c)(b+c-a)(c+a-b)(a+b-c)=0

(y+z)/a^2=(z+x)/b^2=(x+y)/c^2 =k
若k=0 => y=-z; z=-x;x=-y => xyz=-xyz => 2xyz=0 => xyz=0不合 ∴k≠0
=> y+z=ka^2 ; z+x=kb^2 ;x+y=kc^2
得x=0.5(b^2+c^2-a^2)k ;y=0.5(c^2+a^2-b^2)k;z=0.5(a^2+b^2-c^2)k
由xy+z(x+y)=0 => z(x+y)=-xy =kc^2 z 同理 -yz=ka^2x ; -zx=kb^2x
相加 -(xy+yz+zx)=0=k(a^2x+b^2y+c^2z) =>a^2x+b^2y+c^2z=0
=>0.5k[(b^2+c^2-a^2)a^2+(c^2+a^2-b^2)b^2+(a^2+b^2-c^2)c^2]=0
=>(b^2+c^2-a^2)a^2+(c^2+a^2-b^2)b^2+(a^2+b^2-c^2)c^2=0
=>展开 2a^2b^2+2b^2c^2+2c^2a^2 -a^4 -b^4 -c^4=0
但 (a+b+c)(a+b-c)(a+c-b)(b+c-a)=[(a+b)^2 -c^2][c^2-(a-b)^2] 展开
=2a^2b^2+2b^2c^2+2c^2a^2 -a^4 -b^4 -c^4=0