Choose uniformly at random a permutation $\displaystyle \pi \in S_n$. Let $\displaystyle C_\pi = \{\emptyset\} \cup \left\{\pi\left(\{1,...,i\}\right) \ : \ 1 \leq i \leq n\right\}$
And define
$\displaystyle X = |\mathcal{F} \cap C_\pi|$
$\displaystyle X_A = \left\{\begin{array}{cc} 1 & \ A \in C_\pi \\ 0 & \ \text{otherwise} \end{array}\right. \ \forall A \subset [n]$.
It immediately follows that $\displaystyle X = \sum_{A \in \mathcal{F}} X_A$. However, from the way we defined $\displaystyle C_\pi$ we also have that for any set A, it contains exactly one set of size $\displaystyle |A|$, distributed uniformly among all sets of size $\displaystyle |A|$. Thus we have that $\displaystyle \mathbb{E}\left[X_A\right] = \frac{1}{\binom{n}{|A|}}$ and so from linearity of expectation, $\displaystyle \mathbb{E}[X] = \sum_{A \in \mathcal{F}}\mathbb{E}\left[X_A\right] = \sum_{A \in \mathcal{F}}\frac{1}{\binom{n}{|A|}}$
However, since $\displaystyle C_\pi$ is a chain, we must have that $\displaystyle X=\left|C_\pi \cap \mathcal{F}\right| \leq 1,$ therefore
$\displaystyle 1 \geq \mathbb{E}[X] = \sum_{A \in \mathcal{F}}\frac{1}{\binom{n}{|A|}}$
The conclusion follows from the fact that $\displaystyle \binom{n}{x}$ is maximized when $\displaystyle x = \frac{n}{2}$