I have to solve this using the substitution method. Floor functions cannot be skipped.IH: Assume that $T(k) \geq  ck\log(k) $ for all $k \leq n$, where c is a constant.IS: Must prove $T(k) \geq  ck\log(k) $Proof:\begin{align*}T(n) &= 2T(\left \lfloor{k/2}\right \rfloor ) + k\\     &\geq 2 c\left \lfloor{k/2}\right \rfloor\log(\left \lfloor{k/2}\right \rfloor) +k\end{align*}We now have two cases: $k$ is even and $k$ is odd. It is easy to prove this case when $k$ is even, however I run into trouble when $k$ is odd.When k is odd, $\lfloor k/2\rfloor = (k-1)/2$, which gives\begin{align*}T(n)&\geq 2 c\tfrac{k-1}2\log(\tfrac{k-1}2) +k\\&=  c(k-1)\big(\log(k-1)-1\big) +k \qquad\qquad\text{(Since $\log2 = 1$)}\\&=  c\big(k\log (k-1) - k -\log(k-1) + 1\big) +k\\&=  ck\log (k-1) - ck -c\log(k-1) + c +k\\&=  ck\log k  + ck\log\big(1-\tfrac{1}k\big) - ck -c\log(k-1) + c +k\,,\end{align*}where the last equality is because $\log(a-b) = \log\big(a\big(1-\tfrac{b}a\big)\big)$.And here is where I am stuck. I have gotten the $ck\log k$, but cannot find the $c$ that will let me finish the problem. Any help would be appreciated. 
Show that $T(n) = 2T(\lfloor n/2\rfloor) + n$ is $\Omega(n\log n)$ using substitution
recurrence relation;induction;divide and conquer
One way to go around the difficulties is as follows.First, you prove that $T(n) = \Omega(n \log n)$ when $n = 2^k$.Second, you prove that $T(n)$ is monotone (given monotone base cases).Third, you finish the proof as follows. Let $m \leq n < 2m$, where $m = 2^k$. Then$$ T(n) \geq T(m) = \Omega(m \log m) = \Omega(n \log n).$$Details left to you.