Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
Here is my initial question:
If $p$ is a prime number and $k$ is a positive integer, is it true that $$\sigma_1(p^k) > 1 + k (\sqrt{p})^{1+k}?$$
MY ATTEMPT
I dimly recall having been able to prove this assertion last February $6$, $2011$ using the Arithmetic Mean-Geometric Mean Inequality. Indeed, we have
$$\sigma_1(p^k) = \sum_{j=0}^{k}{p^j} > (k+1) \cdot \Bigg(\prod_{j=0}^{k}{p^j}\Bigg)^{\frac{1}{k+1}} = (k + 1) \cdot \Bigg(p^{\displaystyle\sum_{j=0}^{k+1}{j}}\Bigg)^{\frac{1}{k+1}}$$ $$= (k + 1) \cdot \Bigg(p^{\displaystyle\frac{k(k+1)}{2}}\Bigg)^{\frac{1}{k+1}} = (k + 1) \cdot p^{\frac{k}{2}} = (k + 1) \cdot (\sqrt{p})^k,$$
which is weaker than
$$\sigma_1(p^k) > 1 + k (\sqrt{p})^{1+k},$$
as this WolframAlpha computational verification shows.
Alas, I have already forgotten the proof. I think I did it by considering
$$\sigma_1(p^k) = 1 + \sum_{j=1}^{k}{p^j}$$
and then applying the Arithmetic Mean-Geometric Mean Inequality to
$$\sum_{j=1}^{k}{p^j}.$$
I obtain
$$\sum_{j=1}^{k}{p^j} > k \cdot \Bigg(\prod_{j=1}^{k}{p^j}\Bigg)^{\frac{1}{k}}$$ $$= k \cdot \Bigg(p^{\displaystyle\sum_{j=1}^{k}{j}}\Bigg)^{\frac{1}{k}}$$ $$= k \cdot \Bigg(p^{\displaystyle\frac{k(k+1)}{2}}\Bigg)^{\frac{1}{k}}$$ $$= k \cdot (\sqrt{p})^{k+1}.$$
Thus, we have
$$\sigma_1(p^k) > 1 + k \cdot (\sqrt{p})^{1+k},$$
as desired.
However, this seems to result in a contradiction when $k=1$, as $$\sigma_1(p^k) = \sigma_1(p^1) = \sigma_1(p) = p + 1 \not\gt 1 + 1 \cdot (\sqrt{p})^{1+1} = 1 + k \cdot (\sqrt{p})^{1+k}.$$
Here is my final question:
Where is the error in the proof?
The correct inequality to consider is
$$\sigma_1(p^k) \geq 1 + k (\sqrt{p})^{1+k} \tag{1}$$
rather than
$$\sigma_1(p^k) > 1 + k (\sqrt{p})^{1+k}, \tag{2}$$
as Inequality $(1)$ "fixes" the contradiction at $k=1$, while still retaining the "gist" of Inequality $(2)$.