(x²+y²+1)-2(x+y-1)=(x²-2xy+1)+(y²-2y+1)+1=(x-1)²+(y-1)²+1平方大于等于0所以(x-1)²+(y-1)²+1≥1＞0所以(x²+y²+1)-2(x+y-1)＞0x²+y²+1＞2(x+y-1)...
(x²+y²+1)-2(x+y-1)=(x²-2xy+1)+(y²-2y+1)+1=(x-1)²+(y-1)²+1平方大于等于0所以(x-1)²+(y-1)²+1≥1＞0所以(x²+y²+1)-2(x+y-1)＞0x²+y²+1＞2(x+y-1)...