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

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

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 $b$).

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?

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

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