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

设 (y+z)/a^2=(z+x)/b^2=(x+y)/c^2=t
即 (y+z)/t=a^2 (z+x)/t=b^2 (x+y)/t=c^2
ab=(y+z)(z+x)/t=z/t
a^2+b^2-c^2=(y+z+z+x-x-y)/t=2z/t
(a+b+c)(b+c-a)(c+a-b)(a+b-c)=((a+b)^2-c^2)(b+c-a)(c+a-b)
=(a^2+b^2-c^2+2ab)(c^2-a^2-b^2+2ab)
=(2ab+(a^2+b^2-c^2))(2ab-(a^2+b^2-c^2))
=(2ab)^2-(a^2+b^2-c^2)^2
=(2z/t)^2-(2z/t)^2
=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^...