Without knowing what step loses you, it is hard to know where to start. The third step is badly worded and highly condensed.
The proof starts by saying that the arithmetic mean and geometric mean are equal if the number of items being averaged is one. Clear.
The proof then says that if all the items being averaged are equal, then the arithmetic mean and geometric mean are equal to the items being averaged and so must also be equal to each other. Clear
So the case now being considered involves more than one item to be averaged, and not all of them equal to the arithmetic mean. The proof then says, "you may find." What is meant IN PART is that at least one item of the items being averaged MUST be greater than the arithmetic mean and at least one of the being averaged MUST be less than the arithmetic mean. Wrong modal verb. This is assumed without proof, but it is easy enough to prove by contradiction. (For me, the proof is spoiled by this failure to disclose a necessary lemma.) But what the "may find" ALSO means is that you can rearrange the terms of a sum or a product without any effect on the sum or product, and it places one of the greater items (there must be at least) and one of the lesser items (there must at least one) as the final items being added or multiplied. It is a "without loss of generality" statement.
That was the first place in this proof where I had to stop and ponder. Unless you tell us where you are stuck, I doubt any of us wants to try to rewrite a wiki article.
PS You did not say what your math background is. Do you understand proofs by induction in general?