Another way of stating the problem is "If a triangle has area 1 and its three sides are listed in order of length, what is the smallest possible length of the median side?"

First lets assume the triangle with minimum |AC| is scalene, ie |AB|<|AC|<|BC|.

Find the point A' such that A'BC is an isosceles triangle with base BC by simply translating vertex A parallel to BC.  This new triangle will have |A'B|=|A'C|<|BC|.  

But we can also see |AB|<|A'C|<|AC| which means A'BC now has a shorter length of its medial side than the original triangle ABC.

Thus ABC must be isosceles.  One of the congruent legs will be the median length side (those are sides AB and AC).  Let that length be x and let the angle subtended by the legs be t (that is angle BAC).

The area of the triangle can be formulated as (1/2) * x^2 * sin(t).  Set this equal to the given area of 1 and solve for x: x = sqrt(2/sin(t)).

So to minimize x we need to minimize sqrt(2/sin(t)), which is done by maximizing sin(t).  sin(t) reaches its max value of 1 at t=90 degrees.  

Then triangle ABC is a 45-45-90 isosceles triangle.  This triangle has area 1 precisely when |AB|=|AC|=sqrt(2) and |BC|=2.

Therefore the minimum length of |AC| is sqrt(2) units.