Let $A(n)$ be the number of primes less than $n$, divided by $n$ (so for example, $A(n) \leq 1$, as there cannot be more primes less than $n$ as there are integers less than $n$). Suppose that $n$ is a positive multiple of the positive integer $q$. Show that $$A(n) \leq \frac{q+(n-q)\frac{\phi({q})}{q}}{n}$$
This is what I have done.
First note that the above is equivalent to showing that (no. of primes $<n$) $\leq {q+(n-q)\frac{\phi({q})}{q}}$
Let $n=kq$
Therefore (no. of primes $<n$) $=$ (no. of primes $<kq$) $\leq \phi(kq)$ (remember $1$ is not prime)
$\phi(kq)=\phi(k)\phi(q) \leq (k-1)\phi(q)$ as $\phi(k) \leq (k-1)$.
$(k-1)\phi(q) < q+(k-1)\phi(q)=q+(qk-q)\frac{\phi(q)}{q}=q+(n-q)\frac{\phi(q)}{q}$ as required.
The reason I'm worried about this is that I have actually established a stricter inequality than in the question. However, I can't see anything wrong with this. Any reassurance or criticism would be much appreciated.