Question

已知a.b.c.d都是实数.且a2+b2=1.c2+d2=1.求证|ac+bd|小于等于1

急求,帮帮忙

Answer 1

证明：a�0�5+b�0�5=1①
c�0�5+d�0�5=1②
①+②=a�0�5+c�0�5+b�0�5+d�0�5=2.
等式两边同时减去(2ac+2bd)得：
a�0�5+b�0�5-2ac+b�0�5+d�0�5-2bd=2-2ac-2bd
(a-c)�0�5+(b-d)�0�5=2(1-ac-bd)≥0
∴1-ac-bd≥0,即ac+bd≤1。