For any integer $n>2$, there exist $a_1,a_2,\cdots,a_n>0$ that satisfies both$$\tag1a_1+a_2+\cdots+a_n=\frac1{a_1}+\frac1{a_2}+\dots+\frac1{a_n}$$and\begin{equation}\tag2\frac{1}{a_{1}+n-1}+\frac{1}{a_{2}+n-1}+\dots+\frac{1}{a_{n}+n-1}<1\end{equation}
Proof:
We will prove that
If $a_1,\dots,a_n$ not all equal to 1 and satisfy (1) but not (2), then $\frac1{a_1},\dots,\frac1{a_n}$ satisfy both (1) and (2).
It is easy to see that $a_1,\dots,a_n$ satisfy (1) implies $\frac1{a_1},\dots,\frac1{a_n}$ satisfy (1).
It remains to prove that $a_1,\dots,a_n$ don't satisfy (2) implies $\frac1{a_1},\dots,\frac1{a_n}$ satisfy (2).
\begin{align}
& \frac{1}{a_{i}+n-1}+\frac{1}{\frac{1}{a_{i}}+n-1} \\
={}& \frac{1}{a_{i}+n-1}+\frac{a_{i}}{(n-1) a_{i}+1} \\
={}& \frac{a_{i}^{2}+2(n-1) a_{i}+1}{(n-1) a_{i}^{2}+\left(n^{2}-2 n+2\right) a_{i}+n-1} \\
={}& \frac{2\left[a_{i}^{2}+2(n-1) a_{i}+1\right]}{n\left[a_{i}^{2}+2(n-1) a_{i}+1\right]+(n-2)\left(a_{i}-1\right)^{2}} \\
⩽{}& \frac{2}{n}\quad(i=1,2, \dots, n)
\end{align}
Therefore,
$$
\frac{1}{a_{i}+n-1}+\frac{1}{\frac{1}{a_{i}}+n-1}⩽\frac{2}{n}\quad(i=1,2, \dots, n)
$$
equality is attained $⇔(n-2)(a_i-1)^2=0⇔a_i=1$, since $n>2$.
Since not all $a_i$ equal to 1, not all $n$ inequalities attain equality.Adding up these $n$ inequalities gives
\begin{equation}
\sum_{i=1}^n\frac{1}{a_{i}+n-1}+\sum_{i=1}^n\frac{1}{\frac{1}{a_{i}}+n-1}<2
\end{equation}
If $a_1,\dots,a_n$ don't satisfy (2), the first sum $⩾1$, so the second sum $<1$, so $\frac1{a_1},\dots,\frac1{a_n}$ satisfy (2).
My question:
Can you give an example of $a_1,\dots,a_n>0$ that satisfy (1) but not (2)?
I doubt that if $a_1,\dots,a_n>0$ satisfy (1) then they always satisfy (2)?
Best Answer
I think you need to adjust the statement of the thing you want to prove, since the existence of $\{ a_1, \dots, a_n \}$ is proven by simply setting $a_i = 1, \forall i$. If you want to see a sketch proof of how (1) implies (2), use, as suggested in the comments, the method of Lagrange multipliers. With $b_i := a_i - \frac{1}{a_i}$ this is formulated as
\begin{equation} L(b_i, \lambda) = \sum_{i=1}^n \frac{1}{(n-1) + \frac{b_i + \sqrt{b_i^2 + 4}}{2}} - \lambda \sum_{i=1}^n b_i, \qquad \sum_{i=0}^n b_i = 0 \end{equation}
Computation leads to
\begin{equation} \lambda = -\frac{1}{\sqrt{b_i^2 + 4}} \frac{\frac{b_i + \sqrt{b_i^2 + 4}}{2}}{\left((n-1)+\frac{b_i + \sqrt{b_i^2 + 4}}{2} \right)^2} \end{equation}
Without solving one can see that $b_i(\lambda)=b_j(\lambda), \forall i,j$. Implicitly solving and plugging into the constraint
\begin{equation} 0 = \sum_{i=1}^n b_i = \sum_{i=1}^n b_i(\lambda) = n \cdot b(\lambda) \end{equation}
Hence $b_i = 0 $. So the extremum is attained at $L(0,\lambda) = \sum_{i=1}^n \frac{1}{n-1 + 1 } = n \cdot \frac{1}{n} = 1$. It is a sketch, as of course one could make the fact that this is a global maximum, rigorous. Still, in conclusion, you cannot find $a_1, \dots, a_n > 0$ that satisfies only (1) and not (2).
Edit:
I guess the other direct "analytic" way of seeing that $b_i = 0$ is the maximum, is to define the map $f: \mathbb{R}^{n-1} \to \mathbb{R}$
\begin{equation} (b_1, \dots, b_{n-1}) \mapsto \frac{1}{(n-1) + \frac{-(b_1 + \dots + b_{n-1}) + \sqrt{(b_1 + \dots + b_{n-1})^2 + 4}}{2}} + \sum_{j=1}^{n-1}\frac{1}{(n-1) + \frac{b_j + \sqrt{b_j^2 + 4}}{2}} \end{equation}
which lives directly on the hyperplane $b_1 + \dots + b_n = 0 $ and check that its partial derivatives are all negative, except when at the origin.