Lower bounds on the Threshold function
I don't know how you can get or see the bound of $\sqrt{(t+1)(n-t+1)}$ from the original bound $\sqrt{n(n-\vert (2(t-1) -n+1 \vert )}$ but here is the proof that this bounds are asymptotically equal up to a constant factor:

First see that (I exclude $t = 0$ because the threshold function is always $1$)
$$ n(n-\vert (2(t-1) -n+1 \vert) = \left\lbrace 
\begin{array}{ll} 
n(2t-1) &amp; 1 \leq t \leq n/2+1/2\\
n(2n-2t+1) &amp; n/2+1/2 \leq t \leq n-1
\end{array} \right. $$

Define $f_1(t) = n(2t-1)$, $f_2(t) = n(2n-2t+1)$ and $g(t) = (t+1)(n-t+1)$.

Now you have to calculate the maximal value (according to $t$ within the defined intervalls) of the fractions $f_1(t)/g(t)$, $f_2(t)/g(t)$, $g(t)/f_1(t)$ and $g(t)/f_2(t)$. You can do this with differential calculus or approximation with the help of the graph (with $n$ large enough):

$f_1(t)/g(t) \leq f_1(n/2+1/2)/g(n/2+1/2) \leq \dfrac{n^2}{n^2/4} = 4$

$f_2(t)/g(t) \leq f_2(n/2+1/2)/g(n/2+1/2) \leq \dfrac{n^2}{n^2/4} = 4$

$g(t)/f_1(t) \leq g(1)/f_1(1) = \dfrac{2n}{n} = 2$

$g(t)/f_2(t) \leq g(n-1)/f_2(n-1) = \dfrac{n/2}{n/3} \leq 3/2$

This gives you 
$$n(n-\vert 2(t-1)-n-1\vert) = \Theta((t+1)(n-t+1))$$ 
and also the wanted result 
$$\sqrt{n(n-\vert 2(t-1)-n-1 \vert)} = \Theta(\sqrt{(t+1)(n-t+1)}).$$

Is there an easier way to see/get this result?