Solution:

First note that, for given $E, F$, there is only one point $G$ with the required properties: since $BG = FG$, $G$ must lie on the perpendicular bisector of $BF$, and by the second condition, $G$ lies on the line through the midpoint of $EF$ parallel to $BC$; $G$ must thus be the unique intersection of these two lines. We now give an alternative construction for $G$ from which the result will follow.

Since $BA, CD$ are parallel and equal, we may translate $\triangle CDE$ to give a triangle $\triangle BAE'$. Then $E', A, E$ are collinear, with $E'E = E'A + AE = ED + AE = AD = BC$, and $E'E, BC$ are parallel, so we may also translate $\triangle BCF$ to give $\triangle E'EF'$. Then $F, E, F'$ are collinear. Now $E'F' = BF$, and $\angle E'F'E = \angle BFC = 90^\circ = \angle BFE$; hence, $BE'F'F$ is a rectangle. Let $G'$ be its center. Certainly $BG' = FG'$. Let the line through $G'$ parallel to $BC$ hit $BE', FF'$ at $M, N$ respectively; then symmetry gives $E'M = FN$. However, by translation, $EN = E'M$, so $EN = FN$, and the parallel to $BC$ through $G'$ bisects $EF$. Thus, by the uniqueness of $G$ already proven, $G = G'$, the center of $BE'F'F$.

Now,
$$2 \cdot FG = FE' = \sqrt{BF^2 + BE'^2} > \sqrt{2 \cdot BF \cdot BE'}$$
(by AM-GM, strict because $BF < BC = CD < CE = BE'$)

$$= \sqrt{2[ BFF'E' ]} = \sqrt{2( [BAE'] + [E'EF'] + [ABFE] )}$$
(here brackets denote areas)
$$$$
$$\begin{gathered} = \sqrt{2([CDE] + [BCF] + [ABFE])} = \sqrt{2[ABCD]} \\ = \sqrt{2} \cdot AB = AC. \end{gathered}$$

Solution:

The following solution, due to Philip Sung of Saratoga High School, not only establishes the inequality, but computes the difference.

Let $I$ bisect $\overline{BF}$ and $J$ bisect $\overline{EF}$. Extend $AD$; and draw a parallel to $\overline{FE}$ through $B$; let this meet $AD$ at $H$. Extend $\overrightarrow{BF}$ so it meets $DC$ at $K$.

We will show that $FBHE$ is the image of $FIGJ$ under a scaling with ratio $2$ and center $F$. Obviously, $F$ is taken to $F$, $I$ is taken to $B$, and $J$ to $E$. $G$ is the intersection of $IG$ and $JG$, so it is mapped to the intersection of their images, $\overline{BH}$ is the image of $\overline{IG}$ because it is parallel to $\overline{IG}$ and passes through the image of $I$ (i.e., $B$). Similarly, $\overline{EH}$ is the image of $\overline{JG}$. $H$ lies on both $\overline{EH}$ and $\overline{BH}$, so it is the image of $G$.

$\angle HBF$ is right, as is $\angle ABC$. So they cut off equal angles: $\angle HCA \cong \angle KBC$. In addition, because $\overline{AB} \cong \overline{CB}$ and $\angle HAB \cong \angle KCB$ (they are both right), $\triangle HAB \cong \triangle KCB$ and $\overline{AH} \cong \overline{CK}$.

$FH = 2FG$ because $H$ is the image of $G$ under aforementioned dilation. The problem statement then reduces to:

$$AC < FH.$$

We will show, equivalently, that $AC^2 < FH^2$ ($AC$ and $FH$ are both positive.)

$$\begin{array}{rlrl} FH^2 & = FB^2 + BH^2 & & (\triangle FBH \text{ is right}) \\ & = FB^2 + BA^2 + AH^2 & & (\triangle BAH \text{ is right}) \\ & = FB^2 + BA^2 + CK^2 & & (AH = CK) \\ & = FB^2 + BA^2 + CF^2 + FK^2 & (\triangle CK \text{ is right}) \\ & = (FB^2 + CF^2) + BA^2 + FK^2 & \\ & = BC^2 + BA^2 + FK^2 & & (\triangle BFC \text{ is right}) \\ & = AC^2 + FK^2 & & (\triangle ABC \text{ is right}) \\ & > AC^2 & & (FK \text{ is nonzero}) \\ FH^2 & > AC^2 & & \end{array}$$

and we're done!

$$\therefore$$