Regular language using pumping lemma
You probably misunderstood the question. For starters, the pumping lemma can't be used to prove that a language is regular, it can only be used to prove that a language isn't regular. The Wikipedia page for the pumping lemma gives a counterexample.

Secondly, as already pointed out by @DavidRicherby, this particular language is not regular, and as it turns out, you can use the Pumping Lemma to prove this statement.

Roughly speaking, the proof goes as follows:

Assume for the sake of contradiction that the language is regular, and thus, the Pumping Lemma holds. The "y" can take one of three forms:


y contains all "a"s
y contains all "b"s
y is of the form $a^ib^j$ for some $i,j &gt; 0$.


Now, can you see how to derive the contradiction?