条件可整理为(x+y)²+(y+1)² = 1.于是(x+2y+1)² = (x+y)²+(y+1)²+2(x+y)(y+1) ≤ 2((x+y)²+(y+1)²) = 2.得x+2y+1 ≤ √2,即x+2y ≤ √2-1.可验证当x = 1,y = √2/2-1时等号成立,...