Let $A_iA_jA_k$ be a triangle of maximal area. Let $A'_i$ be the reflection of $A_i$ across the midpoint of the side $A_jA_k$; the points $A'_j$ and $A'_k$ are defined similarly. Note that at most one of these three reflections can be a vertex of the polygon — otherwise, the polygon would have three collinear vertices, contradicting convexity.

We will prove that the triangle $A'_iA'_jA'_k$ covers the polygon $A_1A_2...A_n$. By the preceding, the cover is strict, so $[A'_iA'_jA'_k] > [A_1A_2...A_n]$. As $[A'_iA'_jA'_k] = 4[A_iA_jA_k]$, the conclusion follows.

To prove the covering claim above, suppose, if possible, some vertex $A_\ell$ lies outside the triangle $A'_iA'_jA'_k$. Let $\operatorname{dist}(X, YZ)$ denote the Euclidean distance of the point $X$ to line $YZ$. Note that at least one of the three inequalities below holds:
$$\begin{align*} \operatorname{dist}(A_\ell, A_j A_k) &> \operatorname{dist}(A_i, A_j A_k), \\ \operatorname{dist}(A_\ell, A_k A_i) &> \operatorname{dist}(A_j, A_k A_i), \\ \operatorname{dist}(A_\ell, A_i A_j) &> \operatorname{dist}(A_k, A_i A_j). \end{align*}$$
Hence at least one of the triangles $A_\ell A_j A_k$, $A_\ell A_k A_i$, $A_\ell A_i A_j$ has an area (strictly) greater than that of $A_i A_j A_k$, contradicting the choice of this latter. This ends the proof.