Q:

In Peano arithmetic, can we define inequality using successor?

In Peano arithmetic (first order), we first define natural numbers using a successor function and Peano axioms, then we define addition (and multiplication), and then, we define inequality, as:
$a\leq b\leftrightarrow\exists c\left(a+c=b\right)$ 
Is there any way to define inequality first, directly from the successor function and Peano axioms? (I mean, if I don't need addition for my purpose, why define it?).

A:

In a precise sense, the answer is no. Namely, let $PA_{succ}$ be the set of PA-theorems in the language containing only the symbol for the successor function; then we can show:

There are models of $PA_{succ}$ with no definable linear ordering.

In particular, this means that there is no first-order formula using only successor which PA proves defines a linear ordering.
Specifically, consider the structure (in the language of successor only) $\mathbb{N}+\mathbb{Z}+\mathbb{Z}$. This is a model of $PA_{succ}$ (this takes a bit of work, but isn't hard), but has no definable linear ordering: consider any automorphism swapping the two $\mathbb{Z}$-parts.
(A bit more thought also shows that there is no formula in the language of successor alone which defines a linear ordering in the standard model $\mathbb{N}$; the key ingredient is the proof that $PA_{succ}$ is complete. And in fact thinking along these lines ultimately shows that no model of $PA_{succ}$ has a definable linear ordering.)

