Prove that graph on $n$ vertices has at most $n$ cuts of size one Can you please help me prove the following claim?

Any graph on $n$ vertices has at most $n$ cuts of size one.

I tried to use induction on $n$ (assume the hypothesis holds true for $k-1$ and add one vertex) but I fail to get any strong argument from the induction hypothesis.
I would also appreciate help on the following related claim:

Any graph with $m$ edges has a cut of size at least $1/2 \cdot m$

Definition of cut:

$C \subseteq E(G)$ is cut if there exists $X \subseteq V(G)$ such that C consists of all edges in $G$ that have one end in $X$ and the other outside of $X$. $G$ doesn't have to be connected.


I believe there can't be more than $n - 1$ cuts of size one.  Suppose there are at least $n$ cuts of size one.
Let $E'$ be the set of edges included in a cut of size one, and let $G' = (V, E')$, i.e. the graph $G$ in which we only keep the edges of $E'$.
Since $E' \geq n$, $G'$ has a cycle $v_1v_2 \ldots v_kv_1$.  Now, $G'$ has a cut of size one containing only $v_1v_2$ (which is also a cut of size one in $G$).  Say $v_1 \in X, v_2 \in \overline{X}$.  But $v_1v_2 \ldots v_kv_1$ is a cycle so ...  
End of the hint :)
As for the other problem, induction should work here.  Find a base case, then look at $G' = G-v$ (for some arbitrary vertex $v$) in the induction step.
Here, $G'$ has a cut with partitions $X$ and $\overline{X}$ with at least half of its edges.  Now we need to put back $v$ in either $X$ or $\overline{X}$.  One of them has at least half of the neighbors of $v$, and from then you should be able to make the appropriate choice (and fill in the details). 
