∵a＞0，b＞0，且

1

2a+b

+

1

b+1

=1，

∴a+2b=

(2a+b)+3(b+1)

2

-

3

2

=

(2a+b)+3(b+1)

2

•(

1

2a+b

+

1

b+1

)-

3

2

=

1

2

[1+3+

3(b+1)

2a+b

+

2a+b

b+1

]-

3

2

≥

1

2

(4+2

3(b+1) 2a+b • 2a+b b+1

)-

3

2

=

4+2 3

2

-

3

2

=

2 3 +1

2

．

当且仅当

3(b+1)

2a+b

=

2a+b

b+1

，a＞0，b＞0，且

1

2a+b

+

1

b+1

=1，即b=

3

3

，a=

1

2

+

3

3

时取等号．

∴a+2b的最小值为

2 3 +1

2

．

故答案为

2 3 +1

2

．