Define the weight of a side $XY$ to be the area assigned to it, and define an antipoint of a side of a polygon to be one of the points in the polygon farthest from that side (and consequently forming the triangle with greatest area).

Lemma For any side $XY$, $Z$ is an antipoint if and only if the line $l$ through $Z$ parallel to $XY$ does not go through the interior of the polygon. (Note that this means we can assume $Z$ is a vertex, as we shall do henceforth).

proof: Clearly, if $Z$ is an antipoint $l$ must not go through the interior of the polygon. Now if $l$ does not go through the interior of the polygon, assume there is a point $Z'$ farther away from $XY$ than $Z$. Since the polygon is convex, the point $XZ' \cap l$ is in the interior of the polygon, which is a contradiction. ■

Suppose for the sake of contradiction that the sum of the weights of the sides is less than twice the area of some polygon. Then let $S$ be the non-empty set of all convex polygons for which the sum of the weights is strictly less than twice the area. It is easy to check that no polygon in $S$ can be a triangle, so we may assume all polygons in $S$ have at least 4 sides.

We first prove by contradiction that there is some polygon in $S$ such that all of its sides are parallel to some other side. Suppose the contrary; then consider one of the polygons in $S$ which has the minimal number of sides not parallel to any other side (this exists by the well-ordering principle). Call this polygon $P = A_1A_2\cdots A_n$, and WLOG let $A_nA_1$ be a side which is not parallel to any other side of $P$.

Then let $A_i$ be the unique antipoint of $A_nA_1$, and let $A_u$ and $A_v$ be respective antipoints of $A_{i-1}A_i$ and $A_iA_{i+1}$. Define $X$ to be the point such that $A_uX\parallel A_{i-1}A_i$, $A_vX\parallel A_iA_{i+1}$.

Now consider the set $T \subset P$ of points that are strictly on the same side of $A_uA_v$ as $A_nA_1$. First of all, for any side in $T$, $A_i$ must be its antipoint, since the line through $A_i$ parallel to $A_jA_{j+1}$ does not go through the interior of $P$. Similarly, any vertex in $T$ is not the antipoint of any side.

We now look at the polygon $P' = A_vA_{v+1}\cdots A_{u-1}A_uX$. First of all, it is clear that $P'$ has fewer sides which are not parallel to any other side than $P$. Using $[\cdot]$ to denote area, we have
$$[P'] - [P] = [A_1A_2\cdots A_{v-1}A_vXA_uA_{u+1}\cdots A_n].$$
The weights of the side $A_jA_{j+1}$ is the same in both $P'$ and $P$ for $v \le j < u$, but for $P'$, the sum of the weights of the remaining two sides is $[XA_uA_iA_v]$, as $A_i$ is an antipoint of both $A_uX$ and $A_vX$. Meanwhile, the sum of the weights of remaining sides for $P$ is $[A_1A_2\cdots A_{v-1}A_vA_iA_uA_{u+1}\cdots A_n]$. Hence the difference in the sums of weights of $P'$ and $P$ is
$$[XA_uA_iA_v] - [A_1A_2\cdots A_{v-1}A_vA_iA_uA_{u+1}\cdots A_n] = [A_1A_2\cdots A_{v-1}A_vXA_uA_{u+1}\cdots A_n],$$
the same as the difference in area (and both differences were positive). Therefore, if the sum of weights of $P$ was less than $2[P]$, then certainly the sum of weights of $P'$ must be less than $2[P']$, so that $P' \in S$. However, this contradicts the minimality of the number of non-parallel sides in $P$, so there exists a polygon in $S$ with opposite sides parallel.

Now, we will let $R$ be the non-empty set of all polygons in $S$ with all sides parallel to the opposite side. Note that all polygons in $R$ must have an even number of sides. We will show that there is a parallelogram in $R$.

Suppose not, and that $Q = B_1B_2\cdots B_{2m}$ is one of the polygons in $R$ with the minimal number of sides, and $m \ge 3$. Let $X = B_1B_2 \cap B_{2m-1}B_{2m}$ and $Y = B_{m-1}B_m \cap B_{m+2}B_{m+1}$. Set $Q' = XB_2B_3\cdots B_{m-1}YB_{m+2}\cdots B_{2m}$. We propose that the increase in the sum of weights going from $Q$ to $Q'$ is at most twice the increase in area, so that $Q' \in R$.

To aid us, we will let $h_X$ and $h_Y$ be the respective distances of $X$ and $Y$ from $B_{2m}B_1$ and $B_mB_{m+1}$.

The increase in weight is
$$\begin{align*} & [XB_{m+1}B_1] + [XB_{2m}B_m] + [YB_mB_{2m}] + [YB_{m+1}B_1] - [B_1B_{2m}B_m] - [B_{2m}B_mB_{m+1}] \\ &= [XB_1Y] + [XB_{2m}Y] - [B_1B_{2m}B_m] + [YB_mX] + [YB_{m+1}X] - [B_{2m}B_mB_{m+1}] \\ &= [XB_1B_{2m}] + \frac{h_Y \cdot B_1B_{2m}}{2} + [YB_mB_{m+1}] + \frac{h_X \cdot B_mB_{m+1}}{2} \end{align*}$$
while the increase in area is $[XB_1B_{2m}] + [YB_mB_{m+1}]$. It remains to show that the first expression is at most twice the second, or in other words, to show that
$$\frac{h_Y \cdot B_1 B_{2m}}{2} + \frac{h_X \cdot B_m B_{m+1}}{2} \le [XB_1 B_{2m}] + [Y B_m B_{m+1}] = \frac{h_X \cdot B_1 B_{2m}}{2} + \frac{h_Y \cdot B_m B_{m+1}}{2},$$
which is equivalent to
$$(h_X - h_Y)(B_1 B_{2m} - B_m B_{m+1}) \geq 0$$
Noting that triangles $B_1B_{2m}X$ and $B_{m+1}B_mY$ are similar, we have $h_X/h_Y = B_1B_{2m}/B_mB_{m+1}$, so the above inequality holds.

With the inequality proven, we now know that $Q' \in R$, and yet $Q'$ has fewer sides than $Q$. This contradicts the minimality of the number of sides of $Q$, so there exists a parallelogram in $R$. However, the sum of the weights of a parallelogram clearly equals twice its area, so this contradicts the entire existence of $S$, as desired.