The problem is as is in the title:
Is it possible to improve on the bound $$D(q^k) < \varphi(q^k)$$ if $k > 1$?
Here, $q$ is a prime number and $k$ is a positive integer. The (deficiency) function $D(x)$ is defined as follows:
$$D(x) = 2x – \sigma(x)$$
where $\sigma(x)$ is the classical sum of divisors of $x$.
Of course, $\varphi(x)$ is just the Euler totient function of $x$. Finally, let $I(x)=\sigma(x)/x$ be the abundancy index of $x$.
MY PROOF FOR THE BOUND
Since in general we just have $k \geq 1$, we have
$$\frac{q+1}{q} = I(q) \leq I(q^k)$$
from which we get
$$\frac{D(q^k)}{q^k} = 2 – I(q^k) \leq 2 – \frac{q+1}{q} = \frac{q-1}{q} = \frac{\varphi(q^k)}{q^k},$$
whereby we finally obtain
$$D(q^k) \leq \varphi(q^k).$$
So if $k > 1$, then I just get
$$I(q) < I(q^k),$$
right? Proceeding similarly as before, I obtain
$$D(q^k) < \varphi(q^k).$$
Here is my specific question:
Will it be possible to come up with a tighter upper bound than $\varphi(q^k)$ for $D(q^k)$ when $k > 1$?
Edited (in response to a comment on November 5, 2020)
The reason I need a bound for $D(q^k)$ when $k>1$ is because of this closely related question.
Best Answer
Let $$f(k,q):=\varphi(q^k)-D(q^k)=q^{k-1}(q-1)-\bigg(2q^k-\frac{q^{k+1}-1}{q-1}\bigg)=\frac{q^{k-1}-1}{q-1}$$
Then we have
$$\frac{\partial f(k,q)}{\partial q}=\frac{((q-1)(k-2)-1)q^k+q^2}{(q - 1)^2 q^2}$$ which is non-negative.
So, we have $$f(k,q)\ge f(k,2),$$
i.e. $$\color{red}{D(q^k)\le \varphi(q^k)-(2^{k-1}-1)}$$
Added :
If $k\equiv q\equiv 1\pmod 4$, then we have $$D(q^k)\le \varphi(q^k)-\frac{5^{k-1}-1}{4}$$ and $$D(q^k)\le\varphi(q^k)-\frac{q^4-1}{q-1}$$ So we have $$3−\frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{5^{k-1}-1}{4}\bigg)\lt I(q^k)+I(n^2)$$ and $$3−\frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)\lt I(q^k)+I(n^2)$$ But I think that these are not better lower bounds for $I(q^k)+I(n^2)$ than $3−\dfrac{q−2}{q(q−1)}$.