因为(a2+b2)(c2+d2)-(ac+bd)2
=(a2c2+a2d2+b2c2+b2d2)-(a2c2+b2d2+2abcd)
=b2c2+a2d2-2abcd
=(bc-ad)2≥0
又ad≠bc
所以(bc-ad)2>0
所以(a2+b2)(c2+d2)>(ac+bd)2.
已知ad≠bc,求证:(a2+b2)(c2+d2)>(ac+bd)2.
已知ad≠bc,求证:(a2+b2)(c2+d2)>(ac+bd)2.
数学人气:790 ℃时间:2020-05-02 14:35:52
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