Finding Maximum Value using AM-GM Inequality

Question

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 $\sum_{i=1}^{n-1}(x_i*x_{i+1})$.

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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.

Answer 1

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 ($RMS \ge AM \ge GM \ge HM \ for\ positive\ real\ numbers$).

Answer 2

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.