In this paper, Lemma B.1 states an interesting inequality.
Proposition: Let \(\{a_{k}\}_{k \geq 1}\) be a sequence of positive reals. Then, for any \(m \geq 1\), \[\sum_{k=1}^{m - 1}\left(\sum_{i=1}^{k} a_{i}\right)^{3} \leq \frac{m^{3}}{3} \cdot \sum_{k = 1}^{m - 1} a_{k}^{3}~.\]
The proof of this inequality as stated in the aforementioned paper uses mathematical induction, along with an application of the convexity of the function \(f(t) = t^{3}\). Here, I give a more general inequality, which is stated in the lemma below. This inequality uses a different proof based on the lower Riemann sum, instead of mathematical induction.Lemma: Let \(\{a_{k}\}_{k \geq 1}\) be a sequence of positive reals. Then, for any \(m \geq 1\), and \(p \geq 1\), \[\sum_{k = 1}^{m - 1}\left(\sum_{i = 1}^{k} a_{i}\right)^{p} \leq \frac{m^{p}}{p} \cdot \sum_{k = 1}^{m - 1} a_{k}^{p}~.\]
Proof: When \(k \geq 1\), we can use the convexity of \(f(t) = t^{p}\) to obtain \[\left(\sum_{i = 1}^{k} a_{i}\right)^{p} = k^{p} \cdot \left(\sum_{i = 1}^{k} \frac{a_{i}}{k}\right)^{p} \leq k^{p} \cdot \sum_{i = 1}^{k} \frac{1}{k}a_{i}^{p} = k^{p - 1} \cdot \sum_{i = 1}^{k} a_{i}^{p}~.\] When \(k = 0\), this bound above still holds, and consequently is valid for \(k \geq 0\). Since \(x_{i} > 0\) for all \(i \geq 0\), and \(k \leq m - 1\), \[\sum_{i = 1}^{k} a_{i}^{p} \leq \sum_{i = 1}^{m - 1} a_{i}^{p}~.\] Therefore, \[\sum_{k = 1}^{m - 1}\left(\sum_{i = 1}^{k} a_{i}\right)^{p} \leq \sum_{k = 1}^{m - 1} \left(k^{p - 1} \sum_{i = 1}^{k} a_{i}^{p}\right) \leq \left(\sum_{k = 1}^{m - 1} k^{p - 1}\right) \left(\sum_{i = 1}^{m - 1} a_{i}^{p}\right)~.\] We finally bound the sum of \(k^{p - 1}\) as \[\sum_{k = 1}^{m - 1} k^{p - 1} = m^{p} \cdot \sum_{k = 1}^{m - 1} \left(\frac{k}{m}\right)^{p - 1} \frac{1}{m} \leq m^{p} \cdot \int_{0}^{1} t^{p - 1} dt = \frac{m^{p}}{p}~,\] and this proves the required inequality.
Some followup questions: