Show that some context free languages must contain more that one non-terminal
Consider the language $L=\{a^mb^{m+n}a^n\ |\ m,n\ge 1\}$. From the assumption that there is a single-nonterminal CFG $G$ for $L$, we can derive constraints which eventually lead to a contradiction.

$G$ is given by the (singleton) set $\{S\}$ of nonterminals, set $\{a,b\}$ of terminals, and some number of rules $S\to w_i$ for $w_i\in\{a,b,S\}^*$. There are some obvious conditions:


There must be some $i$ such that $w_i$ contains $a$.
There must be some $j$ such that $w_j$ contains $b$.
There must be some $k$ with $w_k\in\{a,b\}^*$.


Note that some of $i,j,k$ might coincide, but this does not matter. Now suppose that there is some $l$ such that $w_l$ contains more than one $S$, i.e. $w_l=xSySz$ for some $x,y,z$. Then
$S\to xSySz\to xxSySzySz\to\dots\to xxw_jyw_izyw_jz$ which contains a $b\dots a\dots b$ subsequence, i.e. we get a word outside $L$ (once any  additional $S$ which might occur inside $x,y,z$ are replaced with $w_k$).

So the rules in $G$ really can only be of the forms $S\to u_iSv_i$ or $S\to w_i$ with $u_i,v_i,w_i\in\{a,b\}^*$. Since we can only use one of the latter form when building a word, and we may need arbitrarily many $b$, we get 


For some $i$, either $u_i$ or $v_i$ contains a $b$. We will assume wlog that it is $v_i$; the $u_i$ case is symmetric. 


Suppose that there is also a $j$ such that $v_j$ contains $a$; then $S\to u_iSv_i\to u_ju_iSv_iv_j\to u_iu_ju_iSv_iv_jv_i$, which again contains a $b\dots a\dots b$ subsequence and takes us out of $L$. Therefore


$v_i\in b^*$ for all $i$. 


But the words in $L$ all end with an $a$, so


$v_i=\epsilon$ for all $i$.


But then $G$ is a regular grammar and cannot generate a non-regular language like $L$.