How to prove that $L = \{a^n b^m a^n b^m \mid n,m \ge 0\}$ is not a CFL?
Let $N$ be the constant of the Pumping Lemma for CFL.

Let be $z=a^N b^{N+1} a^N b^{N+1}$.

let's analyze some of the possible decompositions of $z$ in $uvwxy$:

(note that $vx$ has always at least one symbol)


$u=\epsilon$, $v=a^*$, $w=a^*$, $x=a^*$, $y=b^{N+1} a^N b^{N+1}$. 
just pumping once we get a string out of $L$.
$u=a^*$, $v=a^*$, $w=a^*$, $x=a^*$, $y=b^{N+1} a^N b^{N+1}$. same as above.
$u=a^*$, $v=a^*$, $w=a^*$, $x=a^*b^*$, $y=b^* a^N b^{N+1}$. same.
$u=a^*$, $v=a^*$, $w=a^*$, $x=b^*$, $y=b^* a^N b^{N+1}$. same.
$u=a^*$, $v=a^*$, $w=a^*b^*$, $x=b^*a^*$, $y=a^* b^{N+1}$. it can't happen because $\mid vwx \mid$ would be $&gt;N$.
$u=a^*$, $v=a^*$, $w=a^*b^{N+1}$, $x=a^*$, $y=a^* b^{N+1}$. it can't happen because $\mid vwx \mid$ would be $&gt;N$.
$u=a^*$, $v=a^*$, $w=b^{N+1}$, $x=a^*$, $y=a^* b^{N+1}$. it can't happen because $\mid vwx \mid$ would be $&gt;N$.
$...$


Anyway the problematic case is 7. because it could be $u=a^k$, $v=a^j$, $w=b^{N+1}$, $x=a^j$, $y=a^kb^{N+1}$ with $k+j=N$, so that pumping doesn't push $z$ out of $L$. But again, it can't happen otherwise we would have $\mid vwx\mid &gt;N$, which is not possible and it is guaranteed by the lemma.