How can I show that if $G$ is a finite group that acts on a set $X$ and $m$ denotes the number of orbits that $$m \ge \frac{|X|}{|G|}\:$$
Is it enough to intuitively say that $\sum|\mbox{Orb}_{G}(x)|=|X|$, thus $m=|X|$ in the case that the order of the group $G$ is one, otherwise it must be greater?
The definition of the orbit of $x\in X$ is: $O_x:=\{g\cdot x, g\in G\}$, and hence $|O_x|\le|G|$ for every $x\in X$. Now, the set of the orbits $\mathcal{O}:=\{O_x,x\in X\}$ forms a partition of $X$; therefore:
$$|X|=\sum_{O_x\in\mathcal{O}}|O_x|\le\sum_{O_x\in\mathcal{O}}|G|=m|G|$$
whence $m\ge\frac{|X|}{|G|}$.
Alternatively. By Burnside's lemma, $m=\frac{1}{|G|}\sum_{g\in G}|\operatorname{Fix}(g)|$, where $\operatorname{Fix}(g):=\{x\in X\mid g\cdot x=x\}$. Since $e\in G$ fixes every $x\in X$, $|\operatorname{Fix}(e)|=|X|$ and then:
\begin{alignat}{1} m &= \frac{1}{|G|}\sum_{g\in G}|\operatorname{Fix}(g)| \\ &= \frac{1}{|G|}\bigl(|\operatorname{Fix}(e)|+\sum_{g\in G\setminus\{e\}}|\operatorname{Fix}(g)|\bigr) \\ &= \frac{1}{|G|}\bigl(|X|+\sum_{g\in G\setminus\{e\}}|\operatorname{Fix}(g)|\bigr) \\ &= \frac{|X|}{|G|}\bigl(1+\frac{1}{|X|}\sum_{g\in G\setminus\{e\}}|\operatorname{Fix}(g)|\bigr) \\ &\ge\frac{|X|}{|G|} \end{alignat}