tough lp inequalities Let $1<p<\infty$.  If possible, find a positive decreasing sequence $w_1>w_2>\cdots$ such that $\lim w_i=0$, and a (uniform) constant $K>0$, such that
\begin{equation*}\sum_{i=1}^nw_ia_i\geq K\left(\sum_{i=1}^na_i^p\right)^{1/p}\end{equation*}
for all $n\in\mathbb{N}$ and all nonnegative and nonincreasing sequences $a_1\geq a_2\geq a_3\geq\cdots\geq a_n\geq 0$.
The above is a simplified version of a more complicated problem I'm looking at.  Any ideas would be much appreciated.  Thanks!
 A: First, observe that it suffices to show 
$$
\sum_{i=1}^n w_ia_i \ge \left( \sum_{i=1}^n a_i^p\right)^{1/p}
$$
for all sequences satisfying your conditions and $a_1=1$.
Now let $(w_i)$ and $(a_i)$ be sequences satisfying the non-negativity and monotonicity properties.
Let $n\ge 1$. Define 
$$
f_n (a_n) = \left(\sum_{i=1}^n w_ia_i\right)^p - \sum_{i=1}^n a_i^p.
$$
Let us study the monotonicity of $a_n\mapsto f(a_n)$. It holds
$$
\begin{split}
f_n'(a_n)& = p \left(\sum_{i=1}^n w_ia_i\right)^{p-1}w_n - p a_n^{p-1}\\
& \ge p \left(n w_na_n\right)^{p-1}w_n - p a_n^{p-1} \\
&= pa_n^{p-1}(n^{p-1} w_n^p-1).
\end{split}
$$
This implies that $a_n\mapsto f(a_n)$ is monotonically increasing if 
$$
w_n \ge n^{-(p-1)/p}.
$$
Now, we will show that the choice 
$$
w_n:=\frac1{n^{\frac{p-1}p}}
$$
is sufficient to prove the inequality.
We do this by induction with respect to $n$. For $n=1$ the inequality is trivially true as $w_1=1$. Suppose the inequality holds for some $n\ge 1$ and all feasible sequences. 
Now let $a_1 \ge a_2 \ge \dots \ge a_{n+1}\ge0 $ be given. Consider again the function $f_{n+1}$ from above with the values of $a_1\dots a_n$ being fixed.
From the induction hypothesis, we obtain that $f_{n+1}(0)\ge0$.
Since by our choice of $w_i$ the function $a \mapsto f_{n+1}(a)$ is monotonically increasing, this implies $f_{n+1}(a_{n+1}) \ge0$, which is equivalent to
$$
\sum_{i=1}^{n+1} w_ia_i \ge \left( \sum_{i=1}^{n+1} a_i^p\right)^{1/p}.
$$
And the claim is proven.

If $p=1$ there cannot exists a sequence $(w_i)$ satisfying all conditions. If $a_i=1$ for all $i$, then the inequality is equivalent to
$$
0 \le \left(\sum_{i=1}^n w_i \right)-n = \sum_{i=1}^n (w_i-1)
$$
for all $n$. If $w_i>0$ and $\lim w_i=0$, then the right-hand side tends to $-\infty$ for $n\to \infty$.
