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Today while reading some text I stumbled upon an inequality which basically boiled down to the validity of the following:

If $0\le a_1\le a_2 \le \dots \le a_n$ then $$\frac{a_1+a_2+\dots+a_n}{a_1+2a_2+\dots+na_n}\le\frac{n}{\frac{n(n+1)}2}.\tag{1}$$ The above can be also rewritten as $$\frac{a_1+a_2+\dots+a_n}{a_1+2a_2+\dots+na_n} \le \frac{n}{1+2+\dots+n},\tag{1'}$$ maybe this looks a bit nicer.

I managed to prove it in some way. (See below - I hope I have not made a mistake there.) But I would still be interested to see whether there are some other approaches.


My attempt. We can see that the above inequality is equivalent to $$(n+1)(a_1+a_2+\dots+a_n) \le 2(a_1+2a_2+\dots+na_n). \tag{2}$$ This inequality already seems to be intuitively clear. (The same number of summands on both sides, but on the right hand side we have larger summands more often.)

But it can be also obtained simply by adding the following two inequalities: \begin{align*} a_1+2a_2+\dots+na_n &\le a_1+2a_2+\dots+na_n\\ na_1+(n-1)a_2+\dots+a_n &\le a_1+2a_2+\dots+na_n \end{align*} where the second inequality is a special case of rearrangement inequality.

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From what we are given, since $a_1,a_2,\ldots,a_{k-1} \leq a_k$ we have $$a_1 + a_2 + \cdots +a_k \leq k a_k$$ Hence, $$\sum_{k=1}^n \left(\sum_{l=1}^k a_l\right) \leq \sum_{k=1}^n k a_k$$ Interchanging the sum on the left hand side gives us $$\sum_{k=1}^n (n+1-k)a_k \leq \sum_{k=1}^n ka_k$$ Hence, $$(n+1) \sum_{k=1}^n a_k \leq 2 \sum_{k=1}^n k a_k$$

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