Making a CFG for a^i b^j c^k such that i+k < 3j
$$
\begin{align*}
S \to &amp; AbBC \mid AbBCc \mid AbBCcc \; \mid \\ 
&amp; aAbBC \mid aAbBCc  \mid aAbbBCcc  \; \mid \\
&amp; aaAbBC \mid aaAbbBCc \mid aaAbbBCcc \\
B \to &amp; bB \mid \epsilon \\
A \to &amp; aaaAb \mid \epsilon \\
C \to &amp; bCccc \mid \epsilon 
\end{align*}
$$

The intuition is as follows: 

The nonterminal $B$ encodes the fact that we can have any number (possibly 0) of extra occurrences of $b$s in addition to those strictly needed to satisfy $i+k &lt; 3j$.

For every $3$ occurences of a $a$s or $c$s we have to "pay" by adding one $b$ to our word.
The nonterminal $A$ (resp. $C$) represents a triplet of $a$s (resp. $c$s) along with the occurrence of $b$ that is used to "pay" for the triplet.

Clearly, our word might have a number of $a$s (resp. $c$s) that is not a multiple of $3$, however there are only $9$ possible cases for the combinations of the remainders $i \bmod 3$ and $k \bmod 3$.
These cases are explicitly handled in productions from the axiom $S$, in which we make sure to add at least $\left\lceil 1+\frac{(i \bmod 3) + (k \bmod 3)}{3} \right\rceil$ additional "b"s to "pay" for the $(i \bmod 3) + (k \bmod 3)$ extra characters (the $+1$ ensures that the "greater than" in the inequality $i+k &lt; 3j$ is satisfied).