# Complexity guess and induction proof

I was trying to prove by induction that
$$T(n) = \begin{cases} 1 &\quad\text{if } n\leq 1\ T\left(\lfloor\frac{n}{2}\rfloor\right) + n &\quad\text{if } n\gt1 \ \end{cases}$$ is $$\Omega(n)$$ implying that $$\exists c>0, \exists m\geq 0\,\,|\,\,T(n) \geq cn \,\,\forall n\geq m$$

Base case : $$T(1) \geq c1 \implies c \leq 1$$

Now we shall assume that $$T(k) = \Omega(k) \implies T(k) \geq ck \,\,\forall k < n$$ and prove that $$T(n) = \Omega(n)$$.
$$T(n) = T(\lfloor{\frac{n}{2}}\rfloor) + n \geq c\lfloor{\frac{n}{2}}\rfloor + n \geq c \frac{n}{2} -1 + n \geq n\left(\frac{c}{2} – \frac{1}{n} + 1\right) \geq^{?} cn\ c \leq 2 – \frac{2}{n}$$
So we have proved that $$T(n) \geq c n$$ in :

1) The base case for $$c \leq 1$$

2) The inductive step for $$c \leq 2 – \frac{2}{n}$$

Yet we have to find a value that satisfies them both for all $$n\geq 1$$, the book suggest such value is $$c = 1$$ which to me is not true since :

$$1 \leq 1\1\leq2 – \frac{2}{n}\implies 1 \leq 0 \text{ for n = 1}$$
My guess would be $$0$$ but is not an acceptable value; So we just say its $$\Omega(n)$$ but for $$n \gt 1$$?Or how can we deal with it?