解：|ac+bd丨≤1则(ac+bd)2≤1(ac)2 +(bd)2+2abcd≤1 又(a2+b2)(c2+d2)=1(ac)2 +(bd)2=1-(bc)2 -(ad)2代入不等式得1-(bc)2 -(ad)2+2abcd≤1整理得 (bc)2 +(ad)2-2abcd≥0(bc-ad)2≥0原等式成立
