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Let $x_1,\dots,x_n$ and $y_1,\dots,y_n$ be two increasing sequences of nonnegative real numbers with $x_i\leq y_i$ for all $i$. Is there a constant $c>0$ (independent of $n$) for which there exists some $r\geq 0$ (possibly dependent on $n$ and the sequences) such that $$\sum_{i: x_i\leq r\leq y_i}y_i+\sum_{i:x_i\geq r} x_i\geq c\sum_{i=1}^n y_i?$$

This is a discrete version of this question. For one thing, if $\sum_{i=1}^n x_i$ is not far from $\sum_{i=1}^n y_i$, we can take $r=0$ and the left-hand side is close to the right-hand side.

Example: $x_i=\frac{1}{(n-i)^2}$ and $y_i=\frac{1}{n-i}$. The right-hand side is about $\log n$. For the left-hand side, if we take $r=\frac{1}{n}$, then the first sum is already $\frac{1}{n}+\frac{1}{n-1}+\dots+\frac{1}{\sqrt{n}}$, which is about $\frac{\log n}{2}$, so we can take $c=\frac12$.

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  • $\begingroup$ If $(x_i)_i$ is an unbounded sequence then for any $r$ we have $\sum_{\{i:x_i\geq r\}}x_i=\infty.$ $\endgroup$ Commented May 21, 2017 at 19:33
  • $\begingroup$ @DanielWainfleet No, the sequences are finite sequences. $\endgroup$ Commented May 21, 2017 at 21:09

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No, there doesn't exist. Suppose for contradiction that there exists $c$. We let $x_i=\dfrac{1}{(n-i)^s}$ and $y_i=\dfrac{1}{n-i}$, where $s$ is a function of $c$ that will be chosen later. The right-hand side is about $\log n$. The second term on the left-hand side is bounded above by some constant depending on $s$, call it $f(s)$. Suppose we set $r$ roughly $\dfrac{1}{t}$. Then the first term on the left-hand side will be about $$\frac{1}{t}+\frac{1}{t-1}+\dots+\frac{1}{t^{1/s}}$$ or about $\dfrac{s-1}{s}\log t$, which is maximized when we set $t=n$. For the inequality to hold we need $$\frac{s-1}{s}\log n+f(s)\geq c\log n$$ for all $n$. But if we choose $s$ so that $\dfrac{s-1}{s}<c$, then this cannot be.

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