Sorry for the late answer. May be you can rewrite your problem in the following form
minimize y
such that log(1 + 1/x) \leq y
which is equivalent to
minimize y
such that exp(-y) + exp(-y-log(x)) \leq 1
The following code is working at least
cvx_begin
variables x y
minimize( y )
exp(-y) + exp(-y - log(x) ) <= 1
cvx_end