x-√(x+1)=√(y+3)-yx+y=√(x+1)+√(y+3)≤√2(x+1+y+3)=√慧谨2(x+y+4) (用到(a+b)^2/2≤a^2+b^2)设友碧辩t=x+y则:t≤√2(t+4)t^2≤2(t+4)(^2表示平方)t^2-2t-8=(t-4)(t+2)≤好缺0-2≤t≤4所以, x+y的最大值为4