$(a_1+2a_2+\cdots+na_n)(a_1^2+\cdots+a_n^2)\geq \frac49(a_1+\cdots+a_n)^3$ for non-negative real $a_i$ For any integer $n$ and any nonnegative real numbers $a_1,\ldots,a_n$ we have 
$$(a_1+2a_2+\cdots+na_n)(a_1^2+\cdots+a_n^2)\geq \frac49(a_1+\cdots+a_n)^3$$
It seems to use Holder inequality, but I can't get the $\frac{4}{9}$.
Thanks.

Remark1: @fedja gave an excellent solution by proving a continuous version of the OP which implies the OP (see the link @Martin R pointed out in comment for the OP, which was deleted now). I rewrote fedja's solution for the OP (discrete version).
Remark2: Actually, although @fedja proved a continuous version, his proof can be easily modified to corresponding discrete version (essentially the same).
Proof: For any $b, k > 0$, clearly the following inequality holds:
$$b^2 k + \frac{1}{b}(b - b^3k)^{+} \ge 1$$
where $(x)^{+} = \max(0, x)$. Thus, for any $b>0$, we have 
\begin{align}
\sum a_k &\le \sum \Big(b^2 k + \frac{1}{b}(b - b^3k)^{+}\Big)a_k\\
&= b^2\sum k a_k + \frac{1}{b} \sum (b - b^3k)^{+}a_k\\
&\le b^2\sum k a_k + \frac{1}{b}\sqrt{\sum ((b - b^3k)^{+})^2}\sqrt{\sum a_k^2}\\
&\le b^2\sum k a_k + \frac{1}{b}\sqrt{\frac{1}{3}}\sqrt{\sum a_k^2}\tag{1}
\end{align}
where we have used the following result (the proof is given later)
$$\sum ((b - b^3k)^{+})^2 \le \frac{1}{3}.\tag{2}$$
Since (1) holds for any $b > 0$, by letting
$$b = \sqrt[3]{\frac{1}{2}\sqrt{\frac{1}{3}}\sqrt{\sum a_k^2} \frac{1}{\sum ka_k}},$$
we have
$$\sum a_k \le \frac{3}{2}\sqrt[3]{\frac{2}{3}}\sqrt[3]{\sum k a_k }\sqrt[3]{\sum a_k^2}.$$
The desired result follows. We are done.
$\phantom{2}$
Proof of (2): (by @Martin R) We have 
$$\sum_{k=1}^n ((b-b^3k)^{+})^2 \le \sum_{k=1}^{\lfloor 1/b^2\rfloor} (b-b^3k)^2
\le \int_0^{1/b^2} (b-b^3x)^2 dx = \frac{1}{3}$$
where we have used the fact that $x\mapsto (b-b^3x)^2$ is decreasing on $[0,\frac{1}{b^2}]$.
