(a^2+b^2)(c^2+d^2)>=(ac+bd)^2 (1)<=>a^2c^2+a^2d^2+b^2c^2+b^2d^2>=a^2c^2+2abcd+b^2d^2<=>a^2d^2+b^2c^2>=2abcd<=>(ad)^2-2(ad)(bc)+(bc)^2>=0<=>(ad-bc)^2>=0 (2)不等 展开
(a^2+b^2)(c^2+d^2)>=(ac+bd)^2 (1)<=>a^2c^2+a^2d^2+b^2c^2+b^2d^2>=a^2c^2+2abcd+b^2d^2<=>a^2d^2+b^2c^2>=2abcd<=>(ad)^2-2(ad)(bc)+(bc)^2>=0<=>(ad-bc)^2>=0 (2)不等式(2)显然成立,所以不等式(1)成立. 收起
