设x y为正实数且(√1+x^2+x-1)(√1+y^2+y-1)≤2 则xy的最大值为
问题描述：
设x y为正实数且(√1+x^2+x-1)(√1+y^2+y-1)≤2 则xy的最大值为
答
题目有歧义,建议用标准记号 sqrt{x} 表示x的平方根.1+x方 和1+y方在根号里 (sqrt{1+x^2}+x-1)(sqrt{1+y^2}+y-1)≤2Answer: Max(xy)=1.Denote f(x)=sqrt{1+x^2}+x-1. We compute thatf'(x)=1+x/sqrt(1+x^2).So f(x) is an increasing function on the interval [0,+infinity).Since f(0)=0, we have f(x)>0 for all x>0.(1)Fix x0. According to the given condition, the y maximizing xy must be the maximum y such that f(x0)f(y)