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!

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 $n\to \infty$.