I have a finite sequence of positive numbers $(a_i)_1^n$ for which:
- $a_1>a_n$,
- $a_j\geq a_{j+1}\geq\cdots\geq a_n$ for some $j\in\{2,\ldots,n-1\}$,
- $a_1>a_2>\cdots>a_{j-1}$,
- $a_j\geq a_i$ for all $1\leq i\leq n$.
I conjecture that:
$$(a_n+a_2+\cdots+a_j)\left(\sum_{i=j+1}^{n-1}{\frac{a_i^2}{(a_1+\cdots+a_i)(a_n+a_2+\cdots + a_i)}}+\frac{a_1+a_n}{a_1+\cdots+a_n}\right) \geq (a_n+a_1+\cdots+a_j)\left(\sum_{i=j+1}^{n-1}{\frac{a_i^2}{(a_1+\cdots+a_i)(a_n+a_1+\cdots + a_i)}}+\frac{a_n}{a_1+\cdots+a_n}\right).$$
I have an unappealing brute-force proof when $n\in\{3,4,5\}$ but I can't prove it in general. I have tried calculus to no avail, and it doesn't seem like a good fit for any of the standard inequalities.Does this look even remotely similar to anything already done? I appreciate that it's a rather ugly inequality but some suggestions would be greatly appreciated!
The proof when $n=3$ is outlined below. By condition 2. we know that $j=2$ so that
\begin{align*} (a_3+a_2)\left(\frac{a_1+a_3}{a_1+a_2+a_3}\right)-(a_3+a_1+a_2)\left(\frac{a_3}{a_1+a_2+a_3}\right)=\frac{a_2(a_1-a_3)}{a_1+a_2+a_3}>0 \end{align*} where we have used condition 1, which states that $a_1>a_3$.The proofs when $n=4$ and $n=5$ are likewise, only uglier. When $n=5$ the trick is to find common denominators then simply pair each negative term with some larger positive term. It's horrible, but it works. Perhaps a general proof would involve a similar argument but more formalised?
If a full proof can't be found then I'd be happy for a proof in the special case where $j=2$ and $j=3$.