※ 引述《laiisnotlie ()》之銘言：
: 1.(x+y+z)(x+y)(y+z)(z+x) is not equal to zero
: 2.x^2/(y+z)+y^2/(x+z)+z^2/(x+y)=0
: find out:x^2/yz+y^2/xz+z^2/xy=?
: thank you all
x^2/(y+z) + y^2/(x+z) + z^2/(x+y) = 0
=> x^2(x+z)(x+y) + y^2(y+z)(x+y) + z^2(y+z)(x+z) = 0
(x+y+z)(x+y)(y+z)(z+x) =/= 0
如果其中之一如x = 0,
y =/= -z 且 yz =/= 0
y^2/z + z^2/y = (y^3 + z^3)/yz = 0 yz 不為有限數
=> xyz =/= 0
[x^2/(y+z) + y^2/(x+z) + z^2/(x+y)][1/x + 1/y + 1/z] = 0
[x/(y+1) + y/(x+z) + z/(x+y)] + [x^2/(yz) + y^2/(xz) + z^2/(xy)] = 0
=> [x^2/(yz) + y^2/(xz) + z^2/(xy)] = -[x/(y+z) + y/(x+z) + z/(x+y)]
因為
[x/(y+z) + y/(z+x) + z/(x+y)](x+y)(y+z)(z+x)
= x(z+x)(x+y) + y(x+y)(y+z) + z(y+z)(z+x)
= [x^2(z+x)(x+y) + x(y+z)(z+x)(x+y)]/(x+y+z)
+[y^2(y+z)(x+y) + y(y+z)(x+y)(z+x)]/(x+y+z)
+[z^2(y+z)(x+z) + z(y+z)(x+z)(x+y)]/(x+y+z)
= {(x+y+z)(x+y)(y+z)(z+x)}/(x+y+z)
= (x+y)(y+z)(z+x)
=> x/(y+z) + y/(z+x) + z/(x+y) = 1
所以
x^2/(yz) + y^2/(xz) + z^2/(xy) = -1
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