09/10/26 20:59:26
>>438 (出題元 >>536 から)
(左辺) - (右辺) = x(x-y)(x-z) + y(y-z)(y-x) + z(z-x)(z-y) = F_1,
∴ (x+y)(x+z)(z+x)F_1
= (xy+xz)(x^2 -y^2)(x^2 -z^2) + (yz+yx)(y^2 -z^2)(y^2 -x^2) + (zx+zy)(z^2 -x^2)(z^2 -y^2)
= xy{(x^2 -y^2)(x^2 -z^2) + (y^2 -z^2)(y^2 -x^2)} + cyclic.
= xy(x^2 -y^2)^2 + yz(y^2 -z^2)^2 + zx(z^2 -x^2)^2 ≧ 0,