Is the language $a^i b^j c^k$ with $i+j > k$ context-free?
There is a simple way to obtain a grammar for the language $L_{\geq} = \{a^ib^jc^k : i+j \geq k\}$ given a grammar for the language $L_= = \{a^ib^jc^k : i+j = k\}$. Starting with a grammar for $L_=$, change all rules mentioning $c$ to rules mentioning a new non-terminal $C$, and add the two productions $C\to c \mid \epsilon$.

Concretely, if we start with the following grammar for $L_=$:
$$
\begin{align*}
&amp;S\to aSc \mid T \\
&amp;T\to bTc \mid \epsilon
\end{align*}
$$
then the corresponding grammar for $L_{\geq}$ is
$$
\begin{align*}
&amp;S\to aSC \mid T \\
&amp;T\to bTC \mid \epsilon \\
&amp;C\to c \mid \epsilon
\end{align*}
$$

Of course, this is not quite the restriction we were after. It is possible to modify the grammar for $L_{\geq}$ to a grammar for $L_{&gt;} = \{a^ib^jc^k : i+j &gt; k\}$ by "signalling" within the grammar that at least one $c$ was actually dropped. This requires duplicating some of the non-terminals and rules. Details left to you.

Another modification which will produce $L_{\geq}$ from $L_=$ replaces each $a$ by a non-terminal that generates $a^+$, and each $b$ by a non-terminal that generates $b^+$:
$$
\begin{align*}
&amp;S\to ASc \mid T \\
&amp;T\to BTc \mid \epsilon \\
&amp;A \to Aa \mid a \\
&amp;B \to Bb \mid b
\end{align*}
$$
Again we can use signalling to get $L_&gt;$ rather than $L_{\geq}$. Details left to you.