Given ad ≠ BC, prove: (A2 + B2) (C2 + D2) > (AC + BD) 2

Because (A2 + B2) (C2 + D2) - (AC + BD) 2 = (a2c2 + a2d2 + b2c2 + b2d2) - (a2c2 + b2d2 + 2abcd) = b2c2 + a2d2-2abcd = (BC AD) 2 ≥ 0 and ad ≠ BC, so (BC AD) 2 ＞ 0, so (A2 + B2) (C2 + D2) ＞ (AC + BD) 2

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