用反证法.
假设a²+b²+c²+d²+ab+cd = 1.
则(a+b)²+(c+d)²+(a-d)²+(b+c)² = 2(a²+b²+c²+d²+ab+cd)-2(ad-bc) = 0.
由a,b,c,d都为实数,只有a+b = c+d = a-d = b+c = 0.
解得a = b= c = d = 0,与ad-bc = 1矛盾.
故a²+b²+c²+d²+ab+cd ≠ 1 (实际上有a²+b²+c²+d²+ab+cd > 1).
用反证法.
假设a²+b²+c²+d²+ab+cd = 1.
则(a+b)²+(c+d)²+(a-d)²+(b+c)² = 2(a²+b²+c²+d²+ab+cd)-2(ad-bc) = 0.
由a,b,c,d都为实数,只有a+b = c+d = a-d = b+c = 0.
解得a = b= c = d = 0,与ad-bc = 1矛盾.
故a²+b²+c²+d²+ab+cd ≠ 1 (实际上有a²+b²+c²+d²+ab+cd > 1).