若实数a，b满足a+b2=1，则a2+b2的最小值是．                 �

若实数a，b满足a+b²=1，则a²+b²的最小值是　　　　　　．                       

　．                                                                                                

【考点】配方法的应用；非负数的性质：偶次方．                                         

【分析】由a+b²=1，得出b²=1﹣a，代入得到a²+b²=a²+1﹣a，利用配方法即可求解．                 

【解答】解：∵a+b²=1，                                                                         

∴b²=1﹣a，                                                                                       

∴a²+b²=a²+1﹣a=（a﹣）²+≥，                                                        

∴当a=时，a²+b²有最小值．                                                              

故答案为．                                                                                      

【点评】本题考查了配方法的应用，非负数的性质，将b²=1﹣a代入得到a²+b²=a²+1﹣a是解题的关键．