>>609
(4)
(1-i)(x+iy)(y+iz)(z+ix) = (1-i){-(xyy+yzz+zxx-xyz) +i(xxy+yyz+zzx-xyz)}
= -(x-y)(y-z)(z-x) +i{(x+y)(y+z)(z+x)-4xyz},
絶対値の2乗をとって
 2(xx+yy)(yy+zz)(zz+xx) = {(x-y)(y-z)(z-x)}^2 + {(x+y)(y+z)(z+x) -4xyz}^2,

>>613
 [前スレ.456]
(abc)^2 +aa +bb +cc +2 -2(ab+bc+ca)≧ 0 を使う?
文献[9] 佐藤(訳)、問題3.85改、練習問題1.90(i)