$(a_1+2a_2+\cdots+na_n)(a_1^2+\cdots+a_n^2)\geq \frac49(a_1+\cdots+a_n)^3$ for non-negative real $a_i$

Question

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.

Answer 1

The following is taken from Problems From the Book: 19.22 on AoPS. Instead of going via the corresponding inequality for integrals, I'll show the discrete version directly.

First note that we can assume that $\sum_{i=1}^n i a_i > 0$, otherwise all $a_i$ are zero and the inequality holds trivially.

Now let $a > 0$ be a constant which is determined later, and define $$ b_i = \begin{cases} \frac a2 (3 - a^2 i) & \text{ if } i \le 3/a^2 \, ,\\ 0 & \text{ if } i > 3/a^2 \, . \end{cases} $$ Then $$ 1 \le \frac{a^2}3 i + \frac{2}{3a} b_i \quad \text{for } 1 \le i \le n $$ and $$ \sum_{i=1}^n b_i^2 \le \int_0^{3/a^2} \frac {a^2}{4} (3 - a^2 x)^2 \, dx = \frac 14 \int_0^3 ( 3-y)^2 \, dy = \frac 94 \, . $$

Using the above estimates and the Cauchy-Schwarz inequality we get $$ \begin{align} \sum_{i=1}^n a_i &\le \frac{a^2}3 \sum_{i=1}^n i a_i + \frac{2}{3a} \sum_{i=1}^n a_i b_i \\ &\le \frac{a^2}3 \sum_{i=1}^n i a_i + \frac{2}{3a} \sqrt{ \sum_{i=1}^n a_i^2 \sum_{i=1}^n b_i^2 } \\ &\le \frac{a^2}3 \sum_{i=1}^n i a_i + \frac 1a \sqrt{ \sum_{i=1}^n a_i^2} \,. \end{align} $$

Finally we choose $a$ such that the right-hand side becomes minimal, i.e. $$ a^3 = \frac{3 \sqrt{ \sum_{i=1}^n a_i^2} }{2 \sum_{i=1}^n i a_i} \, . $$ With this choice of $a$ we get $$ \sum_{i=1}^n a_i \le \left( \frac 32 \right)^{2/3} \left(\sum_{i=1}^n i a_i \right)^{1/3} \left(\sum_{i=1}^n a_i^2 \right)^{1/3} $$ or $$ \tag{*} \left( \sum_{i=1}^n a_i \right)^3 \le \frac 94 \sum_{i=1}^n i a_i \sum_{i=1}^n a_i^2 $$ which completes the proof.

Remark: With $a_i = n - i$ we have asymptotically, for $n \to \infty$, $$ \sum_{i=1}^n a_i \sim \frac 12 n^2 \, , \, \sum_{i=1}^n i a_i \sim \frac 16 n^3 \, , \, \sum_{i=1}^n a_i^2 \sim \frac 13 n^3 $$ and that shows that the constant $9/4$ in $(*)$ is the best possible constant which is independent of $n$.

Answer 2

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 $[0,\frac{1}{b^2}]$.