答案:证法1:(分析法)要证(ac bd)2≤(a2 b2)(c2 d2)成立,即证:a2c2 b2d2 2abcd≤a2c2 a2d2 b2c2 b2d2 成立,即证:2abcd≤a2d2 b2c2 成立,即证:0≤a2d2 b2c2-2abcd=(ad-bc)2成立,上式明显成立.故(ac bd)2≤(a2 b2)(c2 d2).证法2:(综合法)因为a2d2 b2c2≥2abcd(重要不等式),所以(ac bd)2=a2c2 b2d2 2abcd≤a2c2 a2d2 b2c2 b2d2=(a2 b2)(c2 d2).证法3:(作差法)因为(a2 b2)(c2 d2)-(ac bd)2(2分)=(a2c2 a2d2 b2c2 b2d2)-(a2c2 b2d2 2abcd)=b2c2 a2d2-2abcd=(bc-ad)2≥0,所以(ac bd)2≤(a2 b2)(c2 d2).