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 $b$).



Coefficient bounds of an inequality
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). 


