Proving that $T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + T(\lfloor n/8 \rfloor) + n$ is $\in O(n)$
First, let me mention that $O(n)$ and $O(cn)$ are exactly the same thing. What you are really after is showing that $T(n) \leq cn$ for all $n$.

Let us aim at a slightly more relaxed goal: showing that $T(n) \leq An$ for some possibly larger constant $A$. For the base case, we need $A \geq c$. For the inductive step, we know that
$$
T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + T(\lfloor n/8 \rfloor) + n \leq A(n/2+n/4+n/8)+1 = (\tfrac{7}{8}A+1)n.
$$
Recall that our goal is to deduce that $T(n) \leq An$. This would follow if $\tfrac{7}{8}A + 1 \leq A$, that is, if $A \geq 8$.

In total, if we take $A = \max(c,8)$ then both the base case and the inductive step go through.



As an aside, you can also use the Akra-Bazzi theorem to directly conclude that $T(n) = O(n)$.



Now let us try to obtain more insight on the recurrence. Let $S(n) = 8n - T(n)$. Then
$$
S(n) = S(\lfloor n/2 \rfloor) + S(\lfloor n/4 \rfloor) + S(\lfloor n/8 \rfloor) + 8n - 8\lfloor n/2 \rfloor - 8\lfloor n/4 \rfloor - 8\lfloor n/8 \rfloor - n.
$$
Since $8(n/a-1) &lt; 8\lfloor n/a \rfloor \leq 8(n/a)$, we see that
$$
S(n) = S(\lfloor n/2 \rfloor) + S(\lfloor n/4 \rfloor) + S(\lfloor n/8 \rfloor) + r(n),
$$
where $0 \leq r(n) &lt; 24$. Applying the Akra-Bazzi theorem, we get $S(n) = O(n^p)$ for some $p &lt; 1$ (the solution to $(1/2)^p + (1/4)^p + (1/8)^p = 1$), and so $T(n) = 8n + O(n^p)$.