The statement is true. In fact, much more general statements are true. If $G$ is a graph with $n$ vertices and $c$ is the cardinality of a minimum edge cut of $G$, then the number of edge cuts of cardinality $c$ is at most $\binom{n}{2}$, and for every half-integer $k \geq 1$, the number of edge cuts containing at most $kc$ edges is bounded above by $2^{2k-1} \binom{n}{2k}.$
The upper bound of $\binom{n}{2}$ on the number of minimum cuts is attributed to Bixby and Dinitz-Karzanov-Lomonosov. The more general bound on the number of approximate minimum cuts is due to Karger (Global min-cuts in RNC, and other ramifications of a simple min-cut algorithm), who also re-proved the $\binom{n}{2}$ bound on minimum cuts. His appealingly simple proof rests on the analysis of a simple "randomized contraction" algorithm. Here we present the proof that the number of minimum cuts is at most $\binom{n}{2}$.
Suppose that $G$ is a multigraph with $n$ vertices, $c>0$ is the number of edges in a minimum cut of $G$, and $A$ is a specific set of $c$ edges whose removal disconnects $G$. Repeatedly perform the following process to obtain a sequence of multigraphs $G = G_0, G_1, \ldots, G_{n-2}$: choose a uniformly random edge of $G_t$ and contract it to obtain $G_{t+1}$. In other words, if $(u,v)$ is the edge chosen in step $t$, then we replace $u$ and $v$ with a single vertex $z$ in $G_{t+1}$, and we replace every edge of $G_t$ having exactly one endpoint in $\{u,v\}$ with a corresponding edge of $G_{t+1}$ with endpoint $z$. (Edges from $u$ to $v$ in $G_t$ are deleted during this step.) Note that $G_{n-2}$ has exactly two vertices $a,b$, these vertices correspond to a partition of $V(G)$ into two nonempty sets $A,B$ (those vertices that were merged together to form $a \in V(G_2)$, and those that were merged together to form $b$), and that the edges of $G_{n-2}$ are in one-to-one correspondence with the edges of the cut separating $A$ from $B$ in $G$. Denote this random cut by $R$.
Now consider a specific cut $C$ of cardinality $c$. We claim that the probability of the event $R=C$ is at least $1 \left/ \binom{n}{2} \right.$, from which it follows immediately that the number of distinct cuts of cardinality $c$ is at most $\binom{n}{2}$. To prove the upper bound on the probability that $R=C$, observe that for all $t = 0,\ldots,n-2$, every vertex of $G_t$ has degree at least $c$. (Otherwise, that vertex of $G_t$ corresponds to a set of vertices in $G$ having fewer than $c$ edges leaving it, contradicting our assumption about the edge connectivity of $G$.) Consequently, the number of edges of $G_t$ is at least $(n-t)c/2$, and the probability that an edge of $C$ is contracted in step $t$, given that no edge of $C$ was previously contracted, is at most $c/|E(G_t)| \leq 2/(n-t)$. Combining these bounds, we find that the probability that no edge of $C$ is ever contracted is bounded below by $\prod_{t=0}^{n-3} \left(1 - \frac{2}{n-t}\right) = \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \frac{1}{3} = \frac{2}{n(n-1)}.$
Updated 4/17/11:
(Originally, this answer contained a different proof of the result below for $k=3$. Not only did the proof not generalize, but it was wrong.)
The maximum number of edges in a strongly-connected digraph with $n \geq k+1$ vertices and no cycles of length at most $k$ is $${\binom{n}{2}} - n(k-2) + \frac{(k+1)(k-2)}{2}.$$
(A digraph where every vertex is reachable from every other vertex by a directed path is called strongly connected.)
Gordon Royle conjectured this bound an gave an example achieving it for $k=3$. For general $k$ and $n$ the bound is attained by the following construction, almost identical to the one provided by Nathann Cohen in the comments:
Let vertices $x_1,x_2,\ldots,x_{n-k+2}$ form a transitive tournament with $x_i \to x_j$ being an edge for all $1 \leq i < j \leq n-k+2$. Now delete the edge $x_1 \to x_{n-k+2}$ and replace it with a path $x_{n-k+2} \to x_{n-k+3} \to \ldots \to x_n \to x_1$. (The vertices $x_{n-k+3},\ldots, x_n$ will have in-degree one and out-degree one in the resulting graph.)
It remains to prove that the above number is a valid upper bound. The proof is by induction on $n$.
Simple counting shows that the bound is valid if $G$ is a directed cycle. It is tight if $G$ is a cycle of length $k+1$. Assume now that $G$ is not a cycle. Then there exist $\emptyset \neq X \subsetneq V(G)$ such that $G|X$ is strongly connected. (For example, one can choose the vertex set of any induced cycle in $G$.) Choose $X$ maximal subject to the above. Let $u \to v_1$ be an edge of $G$ with $u \in X$, $v_1 \not \in X$, and let $P$ be a shortest path in $G$ from $v_1$ to $X$. Let $P=v_1 \to v_2 \to \ldots \to v_l \to w$.
Note that adding to $G|X$ any path starting and ending in $X$ produces a strongly connected digraph. It follows from the choice of $X$ that any non-trivial such path must include all the vertices in $V(G)-X$. In particular, if $l\geq 3$, $v_2,\ldots,v_{l-1}$ have no neighbors in $X$.
Let us further assume that $u$ and $w$ are chosen so that the directed path $Q$ from $w$ to $u$ in $G|X$ is as short as possible. (Perhaps, $w=u$.) Then $V(P) \cup V(Q)$ induces a cycle in $G$, and so $v_1$ and $v_l$ have at least $k-2$ non-neighbors on $V(P) \cup V(Q)$. At least $k-l$ of those non-neighbors are in $X$ if $l\geq 2$. Therefore there are at least $k-2$ non-edges (pairs of non-adjacent vertices) between $X$ and $V(G)-X$ if $l=1$, and at least
$$2(k-l)+(l-2)(k+1) \geq l(k-2)$$
non-edges if $l \geq 2$.
By the induction hypothesis there are at least $|X|(k-2)- \frac{(k+1)(k-2)}{2}$ non-edges between vertices of $X$, and therefore at least
$$(l+|X|)(k-2)- \frac{(k+1)(k-2)}{2}=n(k-2) - \frac{(k+1)(k-2)}{2}$$
non-edges in total, as desired.
Best Answer
Let $N~$ be the number of figures consisting of two triangles sharing an edge, with one of the vertices not on the common edge marked.
Clearly $N=2x$. Alternatively, start with one triangle, mark a vertex, add a second triangle to the opposite side. So $N\le 3t(\Delta_e-1)$. Which is now exact for complete graphs and sharpens your inequality.