Is $L = \{a^jb^ia^{j-i}\mid i,j \ge 0\ , j > i\}$ context-free?
You are trying the wrong way. It is context-free.

(added.) The structure of the language (to guide the grammar construction) is better seen by setting $k = j-i$, so $j=k+i$. Then the language becomes $\{ a^ka^ib^ia^k \mid i\ge 0, k&gt;0 \}$.