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Let $x_1, \ldots ,x_n$ be positive numbers satisfying $|x_{k+1}-x_k| \leq 1$ for any $k$, and $x_1+x_2+ \ldots +x_n > \frac{n(n-1)}{2}$. For $k$ between $1$ and $n$ put

$$ y_k=\frac{x_1+x_2+ \ldots +x_n}{n}-\frac{n-1}{2}+k-1 $$

Then the $y_k$'s satisfy all the conditions satisfied the $x_k$'s, and $\sum_{k=1}^{n} y_k=\sum_{k=1}^{n} x_k$. Now I believe that

$$ \sum_{k=1}^{n} \frac{1}{x_k} \leq \sum_{k=1}^{n} \frac{1}{y_k} $$

Prove or find a counterexample.

Note that the property is true for $n=2$, because

$$ \bigg(\frac{1}{y_1}+\frac{1}{y_2}\bigg)- \bigg(\frac{1}{x_1}+\frac{1}{x_2}\bigg)=\frac{(x_1+x_2)(1-(x_2-x_1)^2)}{4x_1x_2y_1y_2} $$

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up vote 1 down vote accepted

The $x_i$ are more squashed together than the $y_i$.

So, you can transform the $x$s to the $y$s if you successively move two numbers away from each other (always use the two most outer ones that are not in the right $y$ place already).

Now, it is enough to check that the inequality is true for $n=2$ and an arbitrary outwards movement:

Let $y_1=x_1-d$ and $y_2=x_2+d$ with $d<x_1<x_2$ and $x_1$, $x_2$, $y_1$, $y_2$, $d$ positive. We get

$$\begin{multline}\left(\frac 1{y_1} +\frac 1 {y_2}\right)-\left(\frac 1 {x_1}+ \frac 1 {x_2}\right)=\frac{x_2y_2(x_1-y_1)+x_1y_1(x_2-y_2)}{x_1x_2y_1y_2}=\frac{d(x_2y_2-x_1y_1)}{x_1x_2y_1y_2}\\ =\frac{d(x_2(x_2+d)-x_1(x_1-d))}{x_1x_2y_1y_2}= \frac{d(x_2^2-x_1^2)+d^2(x_2+x_1) }{x_1x_2y_1y_2}> 0\end{multline}$$

So, with each step, the expression gets bigger and the final inequality is true. (Note that the final inequality is weak, because there is no guarantee that you need to do a step at all.)

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Indeed, if the $x_i$ are not yet identical to the $y_i$, let $r$ and $s$ be the smallest and largest element in $\lbrace k | x_k \neq y_k \rbrace$. Then necessarily $x_r>y_r$ and $x_s<y_s$, and we replace $(x_r,x_s)$ with $(x_r-d,x_s+d)$ where $d={\sf min}(x_r-y_r,y_s-x_s)$. This will create one more $k$ such that $x_k=y_k$, $k$ will be $r$ or $s$. So everything is finished in at most $n$ steps. – Ewan Delanoy May 27 '12 at 12:34
Nice solution! Thanks. – Ewan Delanoy May 27 '12 at 12:37

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