show that $\lim \inf a_n = \inf a_n$? given that $m = \inf a_n$ and that $\forall n, a_n \not= m $ and $\exists M , |a_n| \leq M $

Prove that $\lim \inf a_n = \inf a_n$ ?

I proved by contradiction that $\inf a_n \not &gt; \lim \inf a_n$

but i don't know how to prove $\inf a_n \not &lt; \lim \inf a_n $

Also if you have a proof not by contradiction, please post it.
We have a bounded sequence which never equals its infimum. This implies that it is always greater than its infimum. 

We must show that $$\lim \inf a_n = \lim_{k \to \infty} \left(\inf_{n &gt; k} a_n\right) = \inf_{n &gt; 0} a_n$$

It suffices to show that $\inf_{n &gt; k} a_n = \inf_{n &gt; 0} a_n$ for arbitrary $k$. Clearly $$\inf_{n &gt; 0} a_n \leq \inf_{n &gt; k} a_n$$ hence it remains to show $$\inf_{n &gt; k} a_n \leq \inf_{n &gt; 0} a_n$$

To prove this, note that if $\inf_{n &gt; k} a_n &gt; \inf_{n &gt; 0} a_n$ we would have $\min(a_1, a_2 \cdots a_k) &lt; \inf_{n &gt; k} a_n $ which would in turn imply that $\min(a_1, a_2 \cdots a_k) = \inf_{n &gt; 0} a_n$. This contradicts the second sentence of this answer.
$inf a_n \lt \lim\inf a_n$ is perfectly possible, unless $a_n$ is a monotone decreasing sequence.  Simple example $a_1=-1, a_n \gt 0,\ n\gt 1$.