Finding the upper bound of the length of a closed walk

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I am having trouble understanding a part of the proof of Lemma 2 (Page 184).

It says the length of the tour is $$ \leq \lceil n^{1/2} \rceil + \triangle(n + \lceil n^{1/2} \rceil) + \sqrt{2} $$ I don't understand how they came up with this inequality. I can understand where the $\sqrt{2} $ comes from, since we have to travel from the final point to the initial point and that is the length of the diagonal. As for the rest, to me somehow it seems that the inequality should be $$ <(\triangle + 1)(n + \lceil n^{1/2} \rceil) + \sqrt{2}$$, where I have used the triangle inequality.

I am having also a hard time simplifying my inequality to something along the lines $c \sqrt{n} $ where $c>0$ whereas the one given in the proof is easier to manipulate.