证明：∵a2+b2=1，x2+y2=1，
∴a2+b2+x2+y2=2，
∵a2+x2≥2ax，b2+y2=2by
∴2ax+2by≤2，
∴ax+by≤1
问题得以证明．
已知a2+b2=1，x2+y2=1，求证ax+by≤1．
已知a2+b2=1，x2+y2=1，求证ax+by≤1．
数学人气：510 ℃时间：2020-04-16 05:09:02
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