a＋b＝(1+sinx,1+cosx)|a+b|^2＝(1+sinx)^2＋(1+cosx)^2＝3＋2(sinx＋cosx)＝3＋2√2sin(x＋π/4)－π/4＜x＋π/4＜3π/4,－√2/2＜sin(x＋π/4)≤1,所以|a+b|^2≤3＋2√2,所以|a+b|≤1＋√2,等号成立时,x＝π/4所以...