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)<=2.(2)Since f(y) is an increasing function in y, with the aid of (1),we conclude that this y must equalize (2), namelyf(x0)f(y)=2.(3)Now we solve y from (3). Observe thatf(x0)f(1/x0)=2.(4).Comparing (3) with (4), we get f(y)=f(1/x0).(5)Since f(y) is strictly monotone (in fact increasing), we infer y=1/x0 from (5).It follows immediately that x0*y=1. In other words, for every x=x0 that possibly maximizes xy,the variable y must be 1/x0.It is easy to check that x=y=1 is such a case. So the maximum value 1 is achievable.This confirms the solution Max{xy}=1.