设x＞0，y＞0，A＝x+y1+x+y

网友回答

∵x＞0，y＞0，
∴x+y+1＞1+x＞0，1+x+y＞1+y＞0
∴x1+x+y＜x1+x
======以下答案可供参考======
供参考答案1:
A=(x+y)/(1+x+y)=1-1/(1+x+y)
B=x/(1+x)+y/(1+y)=（x+2xy+y)/[(1+x)(1+y)
=（x+2xy+y)/(1+x+y+xy)=1-(1-xy)/(1+x+y+xy)
因为(1-xy)/(1+x+y+xy)所以-(1-xy)/(1+x+y+xy)>-1/(1+x+y)
故B>A供参考答案2:
a-b=(x+y/(1+x+y)-(x/(1+x)+y/(1+y))
=(x+y/(1+x+y)-(x+y+2xy)/(1+x+y+xy)
设x+y=m>0,1+x+y=n>0 xy=p>0a-b=m/n-(m+2p)/(n+p)
=[m(n+p)-n(m+2p)]/n(n+p)
=[mp-2np]/n(n+p)
=[(m-2n)p]/n(n+p)
所以a