As noted in the comments, when computing the "big-O" asymptotic form of a function,
you can impose any lower bounds on the input $n$ that you like (for example,
require $n > 34$). And you know that you can multiply an expression by a constant factor
without changing its "big-O" form.
Once you have found that $T(n) \le c\cdot(n+34)\cdot\lg(n+34),$
then you can conclude, for $n > 34,$
$$\begin{eqnarray}
T(n) &\le& c\cdot(n+34)\cdot\lg(n+34) \\
&\le& c\cdot(2n)\cdot\lg(2n) \\
&\le& 2cn\cdot(1 + \lg n) \\
&\le& 2cn\cdot(2 \lg n) = 4c\cdot(n \lg n) = O(n \lg n). \\
\end{eqnarray}$$
You had already found everything up to the last line of that sequence of inequalities.
The key is that you can continue to introduce additional multiplicative constants
as long as it helps simplify your function.
Actually, after some manipulations, you can use the master theorem! Let us see how. First, let us prove the following lemma:
Lemma: The function $T$ is non-decreasing, i.e. $T(n) \leq T(n+1)$ for all $n \in \mathbb{N}$.
Proof. By strong induction on $n \in \mathbb{N}$.
Base cases: For all $0 \leq n \leq 6$, one has $T(n) = 1 = T(n+1)$. Moreover, $T(7) = 1 < 2\,T(0) + 8^{\pi/2} = T(8)$, as $\big\lfloor \frac{8}{\sqrt{2}} \big\rfloor =5$.
Inductive step: Let $n > 7$. The strong induction hypothesis is $T(k) \leq T(k+1)$ for all $0 \leq k < n$. The goal is to prove that $T(n) \leq T(n+1)$.
By definition,
\begin{align}
T(n) &= 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + n^\frac{\pi}{2}
&
T(n+1) &= 2\,T\big(\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor - 5\big) + (n+1)^\frac{\pi}{2}\,.
\end{align}
According to the properties of the floor function, $\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor \leq \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + \big\lfloor \frac{1}{\sqrt{2}} \big\rfloor + 1 = \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + 1$, and $\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor \leq \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + \big\lfloor \frac{1}{\sqrt{2}} \big\rfloor \leq \big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor$, since $\sqrt{2} > 1$.
Therefore, there are only two cases:
- either $\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor = \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor$ and then $T(n) \leq 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + (n+1)^\frac{\pi}{2} = T(n+1)$, where the inequality holds because from $\frac{\pi}{2} > 0$ it follows that $n^\frac{\pi}{2} < (n+1)^\frac{\pi}{2}$ ;
- or $\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor = \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + 1$; we can apply the strong induction hypothesis to $T \big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor- 5 \big)$ because $\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5 < n$ (indeed, $n + 5 > n = \lfloor n \rfloor \geq \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor $ since $\lfloor \cdot \rfloor$ is non-decreasing), so $T \big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor- 5 \big)\leq T \big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor- 5 + 1\big) = T \big(\big\lfloor \frac{n + 1}{\sqrt{2}} \big\rfloor- 5 \big)$ and hence $T(n) \leq 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + (n+1)^\frac{\pi}{2} = T(n+1)$. $\qquad\square$
As $T$ is non-decreasing by the lemma above (and $\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5 < \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor \leq \frac{n}{\sqrt{2}}$), for $n > 7$ one has $T(n) = 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + n^\frac{\pi}{2} \leq 2\,T\big(\frac{n}{\sqrt{2}}\big) + n^\frac{\pi}{2}$. Therefore, if we set
\begin{align}
S(n) =
\begin{cases}
1 &\text{if } n = 0 \\
2\,S\big(\frac{n}{\sqrt{2}}\big) + n^\frac{\pi}{2} & \text{otherwise}
\end{cases}
\end{align}
then $T(n) \leq S(n)$ for all $n \in \mathbb{N}$ and so, for any function $g$, $S(n) \in O(g(n))$ implies $T(n) \in O(g(n))$, i.e. the fact that $S$ grows asymptotically no faster than $g$ implies that $T$ grows asymptotically no faster than $g$.
The point is that we can use the master theorem to find a $g$ such that $S(n) \in O(g(n))$. Using the same notations as in Wikipedia article:
\begin{align}
a &= 2 & b&= \sqrt{2} & c_\text{crit} &= \log_\sqrt{2} 2 = 2 & f(n) &= n^{\pi/2}
\end{align}
thus, $S(n) \in O(n^2)$ by the master theorem (since $\pi/2 < 2 = c_\text{crit}$), and hence $T(n) \in O(n^2)$.
Best Answer
You want to show that there is some $c>0$ such that for all $n$ large enough $T(n) \geqslant cn\lg n$.
I present here a hopefully intuitive inductive approach. We will first check how the inductive step goes.
Suppose that for some $k\geqslant 1$ we have $T(k)\geqslant ck\lg k$.
Can we show that this relation also holds for some other (greater) values $n$?
If we follow the recursive definition of $T$, we'll see that $T(2k)$ and $T(2k+1)$ can both be expressed in terms of $T(k)$.
First, consider the case $n = 2k$. Then $T(n) = 2T(k) + 2k$ and we'll have $T(n) \geqslant c(2k)\lg(k) + 2k$.
It's easy to check that $c\leqslant 1$ implies $c(2k)\lg(k)+2k \geqslant c(2k)\lg(2k)$, regardless of the value of $k$. In this case, $T(n)\geqslant cn\lg n$ as desired.
Now, consider the case $n = 2k+1$. Then $T(n) = 2T(k) + 2k + 1$ and we'll have $T(n)\geqslant c(2k)\lg(k)+(2k+1)$. Under what conditions on $c$ do we have
$$c(2k)\lg(k)+(2k+1) \geqslant c(2k+1)\lg(2k+1)\,\,?$$
That is equivalent to
$$c\leqslant \frac{2k+1}{(2k+1)\lg(2k+1)-2k\lg(k)}.$$
You only really need this to be $\geqslant \epsilon>0$ as $k\to\infty$, and it's easy to show that. For instance, you could show that the RHS converges to $1$ as $k\to\infty$.
That said, you could also show that the minimum is attained at $k=1$ $($with value $\log 2/\log 3\approx 0.63 < 1)$, and this will save us some work in the end with base cases by allowing us to consider base cases right from the get go ($k=1$).
We hence conclude that $c\leqslant \log 2/\log 3$ implies $T(2k+1) \geqslant c(2k+1)\lg(2k+1)$, or $T(n)\geqslant cn\lg n$, once again regardless of the value of $k$.
Combining both cases, we conclude that for $c\leqslant \log 2/\log 3$, $T(k)\geqslant ck\lg k$ implies that both $T(2k)\geqslant c(2k)\lg(2k)$ and $T(2k + 1)\geqslant c(2k + 1)\lg(2k + 1)$.
This will be our inductive step.
To complete the induction, we must now handle the base cases. Our inductive step shows that the relation can be propagated forward from a base case, but not in the traditional way (from $n$ to $n+1$). Rather, our inductive step is a sort of doubling ($n$ to $2n$ and $2n+1$), which in general leaves gaps.
Indeed, for the sake of example, suppose we took a single base case of $k = 7$. Our inductive step would propagate the relation to $n\in\{14, 15\}$, and from these we'd get $n\in\{28,29,30,31\}$, and so on and so forth. It's easy to see that we'd miss many values.
We need a modified induction for our modified inductive step.
Once you've proved the lemma, we can finish the exercise. If we take $j=0$ in the lemma, we'd need only show that $P(1)$ is true. And indeed, it is.
We can calculate the first few values of $T$ by hand: $T(0) = 0$ and $T(1) = 1$. The proposition $P(1)$ is then $T(1) = 1 \geqslant c\lg 1 = 0$, which is obviously true and concludes the induction.
This approach also helps show where tightness was lacking in the answers provided. Looser values of $c$ (meaning values closer to $1$) will require larger values of $k$ in order for the inductive step $k\mapsto 2k+1$ to work.
And, indeed, for $c\geqslant 1$ (including $c=1$) the inductive step does not hold for any $k$.