这不是方程啊？也不是不等式
(a+b-cXa-b+c)+(b+c)F=[a +(b-c)[a-(b-c)]+(b+c)=a2-(b-c)F +(b+c}=a2 +[(b+c)*-(b-c)]= a2 +[(b+c)+(b- c)[(b+c)-(b-c)]=a2+ 2b.2c=a2+ 4bc
