Exercise 65. If $a_1, a_2, \ldots, a_n \in \mathbb{R}^+$ and $s = a_1 + a_2 + \cdots + a_n$, then \begin{equation} \frac{a_1}{s-a_1} + \frac{a_2}{s-a_2} + \cdots + \frac{a_n}{s-a_n} \geq \frac{n}{n-1} . \end{equation}

Now, we have to prove this Inequality using the Rearrangement Inequality. I began by considering assuming (WLOG) that $a_1\leq a_2\leq a_3...\leq a_n$. If $m_i=s-a_i$ then ${1\over m_1}\leq{1\over m_2}.....\leq{1\over m_n}$. Thus we can see that the LHS is maximal. But I am unsure about which permutation of $(a_1,a_2,...a_n)$ should I consider in order to get the RHS.


2 Answers 2



Notice, we can write

$$ \frac{a_i}{s-a_i} = \frac{s}{s-a_i} - 1 $$

Putting $x_i = s - a_i$, notice $\sum x_i = (n-1)s$ and we rewrite the LHS of our inequality as follows

$$ s \left( \frac{ 1 }{x_1} + ... + \frac{1}{x_n} \right) - n $$

Now, taking reciprocals of the HM-AM Inequality we obtain

$$ s \left( \frac{ 1 }{x_1} + ... + \frac{1}{x_n} \right) - n \geq s \frac{n^2}{\sum x_i } - n = s \frac{n^2}{(n-1)s} - n = \frac{n^2}{n-1}-n = \frac{n}{n-1} $$

as desired.


Here is how you can do it with the rearrangement inequality:

As you noticed, the sequences $(a_1,...,a_n)$ and $\left(\frac{1}{s-a_1},...,\frac{1}{s-a_n}\right)$ are ordered in the same way. Thus we have for $i\in\{1,...,n-1\}$: $$ \sum_{k=1}^{n}\frac{a_k}{s-a_k}≥\sum_{k=1}^{n}\frac{a_{k+i}}{s-a_k} $$ Where $a_{n+j}=a_j$ for $j≥1$. By summing up these inequalities we get: $$ (n-1)\sum_{k=1}^{n}\frac{a_k}{s-a_k}=\sum_{i=1}^{n-1}\sum_{k=1}^{n}\frac{a_k}{s-a_k}≥\sum_{i=1}^{n-1}\sum_{k=1}^{n}\frac{a_{k+i}}{s-a_k}=\sum_{k=1}^{n}\frac{1}{s-a_k}\sum_{i=1}^{n-1}a_{k+i}=\sum_{k=1}^{n}\frac{s-a_k}{s-a_k}=n $$


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