设x+2=s，y+1=t，则s+t=x+y+3=4，所以x2x+2+y2y+1=(s−2)2s+(t−1)2t＝(s−4+4s)+(t−2+1t)=(s+t)+(4s+1t)−6＝(4s+1t)−2．因为4s+1t＝14(4s+1t)(s+t)＝14(4ts+st+5)≥94所以x2x+2+y2y+1≥14．故答案为14．...