Can anyone help me get started with the proof attached. I have to use strong induction.
Check the base case (for simplicity make the base case all $\displaystyle n$ up to $\displaystyle 6$), I will assume you have done so.
Now suppose for some $\displaystyle k>6$ that: for all $\displaystyle 0 < r \le k$ that $\displaystyle T(r)<4n$
Now consider:
$\displaystyle T(k+1)=T(\lfloor (k+1)/3 \rfloor)+T(\lfloor (k+1)/5 \rfloor)+T(\lfloor (k+1)/7 \rfloor)+(k+1)$
which by assumption:
.............$\displaystyle <\frac{4(k+1)}{3}+\frac{4(k+1)}{5}+\frac{4(k+1)}{7 }+(k+1)$
Now you should be able to finish this yourself.
CB