y=sinx＋cosx＋2=√2sin(x+π/4)+2
所以y的最小值为2-√2,此时x+π/4=2kπ-π/2,即x=2kπ-π/4
y=sinx＋cosx＋2=√2sin(x+π/4)+2的递减区间为
2kπ+π/2≤x+π/4≤2kπ+3π/2
解得：2kπ+π/4≤x≤2kπ+5π/4