已知正实数a,b满足+=1,则a+b的最小值为( ) A.1       B.2  

已知正实数ab满足+=1,则a+b的最小值为(  )

A1       B2       C4       D2

答案

C【考点】基本不等式.

【分析】利用1与基本不等式的性质即可得出.

【解答】解:正实数ab满足+=1

a+b=a+b=2++2+2=4,当且仅当a=b=2时取等号.

a+b的最小值为4

故选:C

 

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