Coefficient bounds of an inequality

Question

Hello,

Given positive integers $k$ and $n$. Are there upper bounds on coefficients $A$ and $B$ such that they depends only on $k$ (eg., $2 k^k$) and for all non-negative integer sequences $(a_i)_{1}^n, (b_i)_{1}^n$ and non-negative increasing real sequence $(p_i)_{1}^n$, the following inequality holds?

$$ \sum_{i=1}^n b_i \left(\sum_{j=1}^i a_j p_j \right)^k \leq A \sum_{i=1}^n a_i \left(\sum_{j=1}^i a_j p_j \right)^k + B \sum_{i=1}^n b_i \left(\sum_{j=1}^i b_j p_j \right)^k $$

Do you know any result or reference related to the question?

Edit: 11/4 Due to the asymmetry of the left-hand side, we can prove the inequality for $A = k/(k+1)$ and $B = \Theta(k^k)$. Is it possible for the same kind of $A,B$ (up to a constant) such that

\begin{eqnarray*} \sum_{i=1}^n &b_i& (a_1p_1 + \ldots + a_{j-1}p_{j-1} + (a_j + \ldots + a_n) p_j)^k \\ &\leq& A \sum_{i=1}^n a_i (a_1p_1 + \ldots + a_{j-1}p_{j-1} + (a_j + \ldots + a_n) p_j)^k \\ &+& B \sum_{i=1}^n b_i(b_1p_1 + \ldots + b_{j-1}p_{j-1} + (b_j + \ldots + b_n) p_j)^k \end{eqnarray*}

The difficulty is due to the tail $(a_{j+1} + \ldots a_n)p_n$ (idem for $b$).

Answer 1

This is true.

I prefer to denote $q_i=p_i^{-1}$, $\alpha_i=a_ip_i$, $\beta_i=b_ip_i$, $A_i=\sum_{j=1}^i\alpha_j$, $B_i=\sum_{j=1}^i\beta_j$. Now we have to check that $$ \sum_i q_i\beta_iA_i^k\le C\sum_i q_i\alpha_iA_i^k+C\sum_i q_i\beta_iB_i^k $$ This is linear in $q_i$, so we just need to check that $$ \sum_{i=1}^n\beta_iA_i^k\le C\sum_{i=1}^n\alpha_iA_i^k+C\sum_{i=1}^n\beta_iB_i^k $$ for all $n\ge 1$. But $xX^k$ is comparable with $X^{k+1}-(X-x)^{k+1}$ for $0\le x\le X$, so the right hand side is essentially $A_n^{k+1}+B_n^{k+1}$ and the left hand side is dominated by $(A_n+B_n)^{k+1}$. The rest should be clear.

Edit: To cover your second inequality, let's show that the "missing part" $$ \sum_{i=1}^n p_i^k b_i\left(\sum_{j=i}^n a_j\right)^k\le C\sum_{i=1}^n p_i^k a_i\left(\sum_{j=i}^n a_j\right)^k+ C\sum_{i=1}^n p_i^k b_i\left(\sum_{j=i}^n b_j\right)^k $$ holds. By now it shouldn't be surprising that it will suffice to check it for the sequence $p_i$ consisting of several zeroes followed by several ones, in which case it is just exactly the same story as before but written backwards (with summations starting with $n$ and going down).