Upper bound for an expression for distributive lattices Let $L$ be a finite distributive lattice with minimum $0$ and Maximum $1$ and join-irreducible elements $j_1,...,j_l$ and meet irreducible elements $m_1,...,m_l$.
Let $J_L:= \sum\limits_{i=1}^{l}{| [j_i,1]|}$ and $M_L:= \sum\limits_{i=1}^{l}{| [0,m_i]|}$.
Set $X_L$:= $l-|J_L-M_L|$.

Question 1: Is there an easy proof that $X_L >0 $ ? This would prove Frankl's conjecture for distributive lattices (which is already known).
Question 2: What are the distributive lattices with $X_L=l$? Their number starts for $n \geq 3$ with 1,2,1,3,2,7,4. Examples are Boolean lattices.
Question 3: Let $U_n:= \sum\limits_{L \in \mathcal{L}_n}^{}{|J_L-M_L|}$, where $\mathcal{L}_n$ is the set of all distributive lattices on $n$ elements.
$U_n/2$ starts for $n \geq 3$ with 0,0,1,2,6,12,34 (could it be https://oeis.org/A088808 ?). What is $U_n$?

 A: The proposed inequality $X_L>0$ is false. Using Birkhoff's representation theorem, identify $L$ with the lattice of lower ideals in a poset $P$. Then $|P| = \ell$.
The join irreducible elements are the principal lower ideals $(p)$ for $p \in P$. Given a join irreducible element $j=(p)$, the interval $[j,1]$ is the set of ideals $I$ containing $p$. Thus, 
$$J_L = \sum_{p \in P} \#\{ I : I \ni p \} = \sum_{I \in L} \#(I).$$
Similarly, 
$$M_L = \sum_{I \in L} (\ell - \#(I)).$$
So the proposed inequality is
$$\left| \ell \#(L) - 2 \sum_{I \in L} \#(I) \right| < \ell.$$
Let the poset $P$ have $a+b$ elements $x_1$, $x_2$, ..., $x_a$, $y_1$, $y_2$, ..., $y_b$ and the relations that the $x$'s are incomparable, the $y$'s form a chain $y_1 < y_2 < \cdots < y_b$ and $x_i < y_j$ for all $i$, $j$. So the lower ideals are of two types:
(1) Any subset of the $x$'s.
(2) $\{x_1, x_2, \ldots, x_a, y_1, y_2, \ldots, y_k \}$ for $1 \leq k \leq b$.
We have $\ell = a+b$, $|L| = 2^a + b$ and $\sum_{I \in L} \#(I) = (a/2) 2^a + b (a+b/2+1/2)$. So
$$\ell \#(L) - \sum_{I \in L} \#(I) = (a+b)(2^a+b) - (a 2^a + b(2a+b+1))$$
$$=b 2^a - b(a+1).$$
If we choose $a$ and $b$ roughly the same size, then $|b 2^a - b(a+1)|$ will be much larger than $a+b$.
