若x方+Y方=1则xy/x+y+1的最大值为

题目

若x方+Y方=1则xy/x+y+1的最大值为

答案

(1).x²+y²=1.==>(x+y)²-1=2xy,===>(x+y+1)(x+y-1)=2xy.===>xy/(x+y+1)=(x+y-1)/2.故原式z=(x+y-1)/2.===>2z+1=x+y.(2).由均值不等式知,x²+y²≥2xy,===>2(x²+y²)≥x²+2xy+y²=(x+y)².===>√[2(x²+y²)]≥|x+y|≥x+y.又x²+y²=1,故x+y≤√2.===>2z+1≤√2,===>z≤(√2-1)/2.等号仅当x=y=√2/2时取得.故[(xy)/(x+y+1)]max=(√2-1)/2.