The pumping lemma provides a sufficient condition for a language to be non-regular, it is not a necessary condition. This is an example of a language for which the pumping condition holds, so you cannot prove that it is non-regular with (the basic version of) the pumping lemma.
When you see this language definition, you should immediately think of breaking it up:
$$ \begin{align*}
L &= \{ a^i b^j c^k \mid \text{if \(i=1\) then \(j=k\)} \}
= \{ a^i b^j c^k \mid i \ne 1 \vee j=k \} \\
&= \{ a^i b^j c^k \mid i \ne 1 \} \cup \{ a^i b^j c^k \mid j = k \}
= (b^*c^* \cup aaa^*b^*c^*) \cup \{ a^i b^j c^k \mid j = k \} \\
\end{align*} $$
This is taking shape: we've expressed $L$ as the union of a regular language with a language that doesn't look regular (it looks a lot like the classical example of a non-regular language ${b^j c^k \mid j=k}$). We need to work a little more if we can hope to prove that $L$ isn't regular, however: the union of a regular language and a non-regular language can be regular (the union operation can “absorb” the irregularity).
So let's take a step back: what's the problem? In the decomposition above, we have the regular case, with words that begin with no $a$ or with $aa$; and we have the irregular case, where we don't know anything about the initial number of $a$'s. Really, the trouble comes from words that begin with a single $a$. If $L$ is regular, then its intersection $L'$ with the regular language $a(b|c)^*$ would also be regular. Well,
$$L' = L \cap (a(b|c)^*) = \{ a b^j c^k \mid j = k \}$$
If we prove that this language is not regular, then we'll know that $L$ isn't regular either.
To prove that $L'$ is not regular, the pumping lemma works well, just like for $\{b^j c^k \mid j=k\}$. We no longer have this problem with the unbounded number of $a$'s. If $L'$ is regular, then we can find a pumping decomposition for $a b^p c^p$ where $p$ is the pumping length: $a b^p c^p = xyz$ with $y \ne \epsilon$, $|xy| \le p$ and $\forall i, x y^i z \in L'$.
- If $y$ contains the $a$, then $x z$ contains no $a$, which contradicts $x z \in L'$.
- If $y$ contains an unequal number of $b$'s and $c$'s, then $x z$ does not contain an equal number of $b$'s and $c$'s, which contradicts $x z \in L'$.
- If $y$ contains equal numbers of $b$'s and $c$'s and is not $a$, then $y = b^m c^m$ with $m \gt 0$ since $y$ is not empty. Then $x y^2 z$ contains the substring $c b$, which contradicts $x y^2 z \in L'$.
Applying the pumping lemma to $L'$ leads to a contradiction. Therefore the assumption that $L'$ is regular does not hold. Thus $L$ is not regular.
If you are willing to assume that $\{b^j c^k \mid j=k\}$ is not regular, another method to prove that $L'$ is not regular is to read it off an automaton. Suppose that $L'$ is regular; then there is a finite automaton $\mathcal{A}$ that recognizes it. Let $q$ be the state reached by this automaton on the input $a$, and let $\mathcal{A}'$ be this $\mathcal{A}$ with the initial state changed to $q$. If a word is of the form $a w$, then it is accepted by $\mathcal{A}$ if and only if $w$ is accepted by $\mathcal{A}'$. So the language recognized by $\mathcal{A'}$ is $\{w \mid a w \in L' \} = \{b^j c^k \mid j=k\}$. Since this language is known not to be regular, we have a contradiction; the assumption that $L'$ is regular must be false.
