设x2-xy+y2=A
x2-xy+y2=A与x2+xy+y2=1相加可以得到:
2(x2+y2)=1+A (1)
x2-xy+y2=A与x2+xy+y2=1相减得到:
2xy=1-A (2)
(1)+(2)×2得:
2(x2+2xy+y2)=2(x+y)2=3-A≥0
∴A≤3,
(1)-(2)×2得:
2(x-y)2=3A-1≥0,
∴A≥ 13.
综上: 13≤A≤3.
设x2-xy+y2=A
x2-xy+y2=A与x2+xy+y2=1相加可以得到:
2(x2+y2)=1+A (1)
x2-xy+y2=A与x2+xy+y2=1相减得到:
2xy=1-A (2)
(1)+(2)×2得:
2(x2+2xy+y2)=2(x+y)2=3-A≥0
∴A≤3,
(1)-(2)×2得:
2(x-y)2=3A-1≥0,
∴A≥ 13.
综上: 13≤A≤3.