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**The Problem:** Let $a_1,a_2,\dots,a_n$ be positive real numbers, all different. Set $s=\sum_{i=1}^na_i$.

* Prove that $(n-1)\sum_{i=1}^n\dfrac{1}{s-a_i} < \sum_{i=1}^n\dfrac{1}{a_i}$.
* Deduce that $\dfrac{4n}{s} < s\sum_{i=1}^n\dfrac{1}{a_i(s-a_i)} < \dfrac{n}{n-1}\sum_{i=1}^n\dfrac{1}{a_i}$

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From AM >= HM, we have:

\[
  \dfrac{n-1}{a_2+a_3+\ldots+a_n} \leqslant \dfrac{\frac{1}{a_2}+\frac{1}{a_3}+\ldots+\frac{1}{a_n}}{n-1}
\]

The inequality is going to be strict because $a_1,a_2, ... , a_n$ are all different:

\[
  \dfrac{(n-1)^2}{a_2+a_3+\ldots+a_n} < \frac{1}{a_2}+\frac{1}{a_3}+\ldots+\frac{1}{a_n}
\]

This way we prove that any summand of the form $\dfrac{1}{s-a_r}$ is less than the sum of $\dfrac{1}{a_i}$ where $i\neq r$. After applying the same reasoning for $n$ terms, we sum up the inequalities, and we get:

\[
  \sum_{i=1}^n\dfrac{(n-1)^2}{s-a_i} < \sum_{i=1}^n\dfrac{n-1}{a_i}
\]

\[
  (n-1)^2\sum_{i=1}^n\dfrac{1}{s-a_i} < (n-1)\sum_{i=1}^n\dfrac{1}{a_i}
\]

\[
  (n-1)\sum_{i=1}^n\dfrac{1}{s-a_i} < \sum_{i=1}^n\dfrac{1}{a_i}
\]

First part is proven. Now, how do we proceed to the second part?

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Note that:

\[
\sum_{i=1}^n\dfrac{s}{a_i(s-a_i)}=\sum_{i=1}^n\left(\dfrac{a_1}{a_i(s-a_i)}+\dots+\dfrac{\cancel{a_i}}{\cancel{a_i}(s-a_i)}+\dots+\dfrac{a_n}{a_i(s-a_i)}\right)=
\]

\[
=\sum_{i=1}^n\left(\dfrac{\cancel{s-a_i}}{a_i\cancel{(s-a_i)}}+\dfrac{1}{s-a_i}\right)=\sum_{i=1}^n\left(\dfrac{1}{a_i}+\dfrac{1}{s-a_i}\right)=\sum_{i=1}^n\dfrac{1}{a_i}+\sum_{i=1}^n\dfrac{1}{s-a_i}
\]

Therefore:

\[
s\sum_{i=1}^n\dfrac{1}{a_i(s-a_i)}=\sum_{i=1}^n\dfrac{1}{a_i}+\sum_{i=1}^n\dfrac{1}{s-a_i}
\]

From (a) we know that $(n-1)\sum_{i=1}^n\dfrac{1}{s-a_i} < \sum_{i=1}^n\dfrac{1}{a_i}$, therefore:

\[
\sum_{i=1}^n\dfrac{1}{s-a_i} < \dfrac{1}{n-1}\sum_{i=1}^n\dfrac{1}{a_i}
\]

From there, we get:

\[
\sum_{i=1}^n\dfrac{1}{a_i}+\sum_{i=1}^n\dfrac{1}{s-a_i} < \sum_{i=1}^n\dfrac{1}{a_i}+\dfrac{1}{n-1}\sum_{i=1}^n\dfrac{1}{a_i}
\]

\[
\sum_{i=1}^n\dfrac{1}{a_i}+\sum_{i=1}^n\dfrac{1}{s-a_i} < \dfrac{n}{n-1}\sum_{i=1}^n\dfrac{1}{a_i}
\]

\[
\boxed{s\sum_{i=1}^n\dfrac{1}{a_i(s-a_i)} < \dfrac{n}{n-1}\sum_{i=1}^n\dfrac{1}{a_i}}
\]

Wed, 17 Apr 2024 05:51 GMT