设a,b,x,y∈R,且a2+b2=1,x2+y2=1,试证:|ax+by|≤1.
问题描述:
设a,b,x,y∈R,且a2+b2=1,x2+y2=1,试证:|ax+by|≤1.
答
证明:1=(a2+b2)(x2+y2)=a2x2+a2y2+b2x2+b2y2≥a2x2+2aybx+b2y2=(ax+by)2,
故|ax+by|≤1.