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let $a_{i}\ge 0$ and such $$a_{1}+a_{2}+\cdots+a_{n}=1$$ find the maximum of the value $$\left(\sum_{i=1}^{n}ia_{i}\right)\left(\sum_{i=1}^{n}\dfrac{a_{i}}{i}\right)^2$$ I try to use From Pólya-Szegö’s inequality, we have for $0 < m_1 \leqslant u_k \leqslant M_1$ and $0 < m_2 \leqslant v_k \leqslant M_2$, $$\left(\sum u_k^2 \right) \left( \sum v_k^2 \right) \leqslant \frac14 \left( \sqrt{\frac{M_1 M_2}{m_1m_2}} + \sqrt{\frac{m_1 m_2}{M_1 M_2}} \right)^2 \left( \sum u_k v_k\right)^2$$

But I can't it.Thanks

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    $\begingroup$ This seems amenable to Lagrange multipliers: with $S_1=\sum_j j a_j$ and $S_2=\sum_j a_j/j$, you have that $i S_2 + 2 S_1/i$ is independent of $i$. Setting those equal to $\lambda$ and requiring $\sum_j a_j=1$ is then $n+1$ linear equations in $n+1$ unknowns. I see no immediate reason why this system should have some nice solution. $\endgroup$
    – Ian
    Feb 27, 2019 at 2:05
  • $\begingroup$ can you find this inequality when $=?$ $\endgroup$
    – math110
    Feb 27, 2019 at 2:10
  • $\begingroup$ How would this inequality help you deal with the square on the outside of the second sum? $\endgroup$
    – Ian
    Feb 27, 2019 at 2:11
  • $\begingroup$ Note that $$\sqrt{(\sum ia_i)(\sum \frac{a_i}{i})^2}\leq (\sum \sqrt{ia_i})(\sum \frac{a_i}{i})\leq (\sum \frac{i+a_i}{2})(\sum \frac{a_i}{i}) \leq(\frac{n(n+1)}{4}+\frac{1}{2})(\sum \frac{a_i}{i})$$ $$\leq(\frac{n(n+1)}{4}+\frac{1}{2})(\sum a_i) =\frac{n(n+1)}{4}+\frac{1}{2}.$$ Maybe it can works. $\endgroup$
    – guchihe
    Feb 27, 2019 at 2:13
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    $\begingroup$ @Ian for $n>2$ you can get $\dfrac{((n+1)/3)^3}{(n/2)^2}>1$ $\endgroup$
    – Macavity
    Feb 27, 2019 at 4:02

3 Answers 3

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Here's the same answer but via different approach. Let $X$ be an integer-valued random variable such that $$ \Bbb P(X=i)=a_i\ \ \ \text{ for all }\ i=1,2,\ldots,n. $$ We have that $$ \Bbb E\left[\frac1 X\right]=\sum_{i=1}^n\frac{a_i}i,\quad \Bbb E\left[X\right]=\sum_{i=1}^n ia_i, $$ so we need to maximize $\Bbb E\left[\frac1 X\right]^2 \Bbb E\left[X\right]$. Observe that by AM-GM inequality $$ 3\sqrt[3]{\Bbb E\left[\frac{n} X\right] \Bbb E\left[\frac {n} X\right] E\left[2X\right]}\le \Bbb E\left[\frac{n} X\right] +\Bbb E\left[\frac{n} X\right]+\Bbb E\left[2X\right]= \Bbb 2\Bbb E\left[X+\frac n X\right]. $$ Since $1\le X\le n$, we have that $$ 0\le \frac{(X-1)(n-X)}X\implies X+\frac n X\le n+1, $$ with equality holding when $\Bbb P(X\in \{1,n\})=1$. This gives $$ 3\sqrt[3]{2n^2\Bbb E\left[\frac{1} X\right]^2E\left[X\right]}\le 2(n+1), $$ which is equivalent to $$ \Bbb E\left[\frac{1} X\right]^2\Bbb E\left[X\right]\le \frac1{2n^2}\left(\frac{2(n+1)}3\right)^3=\frac{4(n+1)^3}{27n^2}. $$ Equality holds when $$ \Bbb E\left[\frac{n} X\right]=2\Bbb E\left[X\right],\quad \Bbb P(X\in \{1,n\})=1, $$ that is, $a_1=\Bbb P(X=1)=\frac{2n-1}{3(n-1)}$ and $a_n=\Bbb P(X=n)=\frac{n-2}{3(n-1)}$ (with other $a_i=0$ and $ n\ge 2$.)

If $n=1$, $\Bbb E\left[\frac{1} X\right]^2\Bbb E\left[X\right]=1$ holds trivially.

Note: In fact, $X$ does not have to be integer-valued; $\Bbb P(1\le X\le n)$ provides the same upper bound. So, for instance, we have that $$ \left(\int_1^n \frac{f(x)\ dx}x\right)^2\left(\int_1^n xf(x)\ dx\right)\le\frac{4(n+1)^3}{27n^2} $$ for all $f\ge 0$ with $\int_1^n f(x)\ dx=1$, $n\ge 2$.

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Let $f(k_1,\cdots,k_m)$ for distinct positive integers $k_1<k_2<\cdots<k_m$ be the maximum possible value of

$$\left(\sum_{i=1}^m k_ia_i\right)\left(\sum_{i=1}^m \frac{a_i}{k_i}\right)^2$$

across all nonnegative reals $a_1,\cdots,a_m$ with sum $1$.

For $m=1$ the value is simply $1$. For $m\geq 2$ we claim

$$f(k_1,\cdots,k_m)=\max\left(\frac{2(k_1+k_m)^2}{9k_1k_m},1\right).$$

We show this by induction on $m$. Use Lagrange multipliers. Letting $S_1=\sum_{i=1}^m k_ia_i$ and $S_2=\sum_{i=1}^m \frac{a_i}{k_i}$ gives that the derivative with respect to $a_i$ is

$$\frac{2S_1}{k_i}+S_2k_i;$$

the derivative vector of our condition is the all-ones vector, so we require that

$$\frac{2S_1}{k_i}+S_2k_i=\lambda$$

is fixed across all $k_i$. However

$$\frac{2S_1}{k_i}+S_2k_i = \frac{2S_1}{k_j}+S_2k_j \implies S_2(k_i-k_j) = S_1\left(\frac{k_i-k_j}{k_ik_j}\right) \implies 2S_1=k_ik_jS_2,$$

which cannot occur for all pairs $(i,j)$ if $m\geq 3$. Thus one variable is $0$, and we are thus reduced to the case of $k_1,\cdots,k_m$ with one element removed. If $m=2$, the system expands to

$$0=a_1(2k_1-k_2)+a_2(2k_2-k_1),$$

which, in addition to the normalization condition $a_1+a_2=1$, has solution

$$a_1=\frac{2k_2-k_1}{3(k_2-k_1)},\ \ a_2=\frac{k_2-2k_1}{3(k_2-k_1)}.$$

If $k_2\geq 2k_1$ this is a valid point and gives the desired value of $\frac{2(k_1+k_2)^2}{9k_1k_2}$; otherwise one $a_i$ must be $0$ which gives the value of $1$. Thus, the base case is proven. For the inductive step, we have that, per our inductive hypothesis, as one $a_i$ must be $0$, this is $\frac{2(k_1+k_m)^2}{9k_1k_m}$ unless we have chosen to remove $k_1$ or $k_m$, in which case it is $\frac{2(k_1+k_{m-1})^2}{9k_1k_{m-1}}$, $\frac{2(k_2+k_m)^2}{9k_2k_m}$, or $1$. The observation that $f(x)=x+\frac{1}{x}$ is increasing on $x\geq 1$ finishes the proof.

Thus, the maximum possible value is

$$\frac{2(n+1)^2}{9n},$$

at $n\geq 2$ and $1$ otherwise, reached at $a_1=\frac{2n-1}{3(n-1)}, a_2=\cdots=a_{n-1}=0,$ $a_n=\frac{n-2}{3(n-1)}.$

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Alternative solution

Problem: Let $a_i\ge 0, \forall i$ with $a_1 + a_2 + \cdots + a_n = 1$. Prove that the maximum of $$\left(\sum_{i=1}^n i a_i\right)\left(\sum_{i=1}^n \frac{a_i}{i}\right)^2$$ is $\frac{4(n+1)^3}{27n^2}$.

We may use the approach in my answer to this equation: An upper bound of product of two inner products

Consider the optimization problem $$\max_{a_i\ge 0, \forall i; \ \sum_{i=1}^n a_i = 1} \left(\sum_{i=1}^n i a_i\right)\left(\sum_{i=1}^n \frac{a_i}{i}\right)^2.$$ Let $(a_1^\ast, a_2^\ast, \cdots, a_n^\ast)$ be a global maximizer.

We claim that if $1 < k < n$, then $a_k^\ast = 0$. Indeed, if $a_k^\ast > 0$, let $$a_1' = a_1^\ast + (1 - y) a_k^\ast, \quad a_k' = 0, \quad a_n' = a_n^\ast + y a_k^\ast$$ where $\frac{k-1}{n-1} < y < \frac{n}{k}\cdot \frac{k-1}{n-1}$. We have $$a_1' + a_k' + a_n' = a_1^\ast + a_k^\ast + a_n^\ast,$$ and \begin{align} &a_1' + ka_k' + na_n' - (a_1^\ast + ka_k^\ast + na_n^\ast)\\ =\ & (n-1)\left(y - \frac{k-1}{n-1}\right) a_k^\ast\\ >\ & 0, \end{align} and \begin{align} &a_1' + \frac{a_k'}{k} + \frac{a_n'}{n} - \left(a_1^\ast + \frac{a_k^\ast}{k} + \frac{a_n^\ast}{n}\right)\\ =\ & \frac{n-1}{n}\left(\frac{n}{k}\cdot \frac{k-1}{n-1} - y\right)a_k^\ast\\ >\ & 0. \end{align} However, this contradicts the optimality of $(a_1^\ast, a_2^\ast, \cdots, a_n^\ast)$.

Now, since $a_2^\ast = a_3^\ast = \cdots = a_{n-1}^\ast = 0$, we have \begin{align} &\left(\sum_{i=1}^n i a_i^\ast\right)\left(\sum_{i=1}^n \frac{a_i^\ast}{i}\right)^2\\ =\ & (a_1^\ast + n a_n^\ast)\left(a_1^\ast + \frac{a_n^\ast}{n}\right)^2\\ =\ & (1 - a_n^\ast + na_n^\ast)\left(1 - a_n^\ast + \frac{a_n^\ast}{n}\right)^2\\ =\ & \frac{1}{2n^2}[2 + 2(n-1)a_n^\ast][n - (n-1)a_n^\ast]^2\\ \le\ & \frac{1}{2n^2} \left(\frac{2 + 2(n-1)a_n^\ast + n - (n-1)a_n^\ast + n - (n-1)a_n^\ast}{3}\right)^3\\ =\ & \frac{4(n+1)^3}{27n^2} \end{align} with equality if $a_n^\ast = \frac{n-2}{3(n-1)}$, where we have used AM-GM inequality.

Thus, the desired maximum is $\frac{4(n+1)^3}{27n^2}$ at $a_1 = \frac{2n-1}{3(n-1)}, a_2 = a_3 = \cdots = a_{n-1} = 0, a_n = \frac{n-2}{3(n-1)}$.

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