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Let us have a set of natural numbers $S=\{x_1,x_2,...,x_n\}$ where $n≥4$, $n$ is even, such that all $(x_i\in S)≥0$ and $\sum_{i=1}^nx_i=1$.

Find the maximum value of $\sum_{i=1}^{n-1}(x_i*x_{i+1})$.

My approach to the problem:

We have $x_1,x_2,...,x_n≥0$. Since $x_1+x_2+...+x_n=1$, it implies that any $x_i>1$ is not possible. Further, even if one $x_i=1$, then all other have to be zero.
Therefore $0≤x_1,x_2,...,x_n≤1$. Using AM$≥$GM, we have $\frac{x_1^2+x_2^2}{2}≥x_1x_2$ with equality$\iff x_1=x_2$. Now to find $\sum_{i=1}^{n-1}(x_i*x_{i+1})$, we add respective inequalities.$$\frac{x_1^2+2(x_2^2+x_3^2+...+x_{n-1}^2)+x_n}{2}≥x_1x_2+x_2x_3+...+x_{n-1}x_n$$ with equality holding if and only if $x_1=x_2=...=x_n$. It gives $RHS_{max}=\frac{n-1}{n^2}$.

Kindly correct me if there is a mistake. Thank you.
20DPCO190 Amanul Haque
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    It seems you really didn't mean natural numbers were defining $S$, but non-negative real ones. Further, $x_1=x_2=\frac12, x_{n>2}=0$ does a much better job of maximising what you want. – Macavity Mar 30 '24 at 01:57
  • ... or any other pattern with for some $j: 1<j<n$ where $x_j=\dfrac12$ and $x_{j-1}+x_{j+1}=\dfrac12$ and the rest $0$, also giving $\sum\limits_{i=1}^{n-1}x_i,x_{i+1} =\dfrac14$. – Henry Mar 30 '24 at 02:00
  • Related, but not quite the same: https://math.stackexchange.com/questions/252103/maximize-x-1x-2x-2x-3-cdotsx-nx-1 – Henry Mar 30 '24 at 02:11

2 Answers2

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Here is a way to use AM-GM to find this maximum:

$$\sum_{i=1}^{n-1}x_ix_{i+1} \leqslant(x_1+x_3+...)(x_2+x_4+...)\leqslant \frac{(x_1+x_2+x_3+x_4+\cdots)^2}4=\frac14$$ Equality is possible when any two adjacent $x_i, x_{i+1}$ are $\frac12$ and the rest $0$, so this gives the maximum.

Macavity
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  • +1 though there are some other cases where the sum is $\frac14$ too. – Henry Mar 30 '24 at 02:15
  • @Henry Yes, saw your comment, those cases also give the maximum. Seems your linked post is almost identical, it doesn't seem to matter if $x_nx_1$ term is not there (but for more equality cases), or that $n$ is even/odd. – Macavity Mar 30 '24 at 02:22
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    The only substantial difference with the other question seems to be when $n<4$. – Henry Mar 30 '24 at 02:24
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Yes, the maximum value of the sum is $\frac{n-1}{n^2}$.

But the inequality which you have used while solving the problem is not AM-GM inequality. It is RMS-GM inequality ($RMS \ge AM \ge GM \ge HM \ for\ positive\ real\ numbers$).

Sparsh Gupta
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