Does this inequality hold? $\sum a_i=\sum\frac1{a_i}$ then $\sum\frac{1}{a_i+n-1}⩽1$

Question

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)?

Answer 1

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.

Answer 2

Fact 1. Let $n\ge 2$. Let $a_1, a_2, \cdots a_n > 0$ with $a_1+a_2+\cdots+a_n=\frac1{a_1}+\frac1{a_2}+\dots+\frac1{a_n}$. Then $$\frac{1}{a_{1}+n-1}+\frac{1}{a_{2}+n-1}+\dots+\frac{1}{a_{n}+n-1} \le 1.$$

Proof.

Let $x_i = a_i - \frac{1}{a_i}, \forall i$. Then $a_i = \frac12x_i + \frac12\sqrt{x_i^2 + 4}, \forall i$.

The condition becomes $x_i \in \mathbb{R}, \forall i$ with $x_1 + x_2 + \cdots + x_n = 0$. We need to prove that $$\sum_{i=1}^n \frac{1}{\frac12x_i + \frac12\sqrt{x_i^2 + 4} + n - 1} \le 1. \tag{1}$$

WLOG, assume that $x_1 \le x_2 \le \cdots \le x_n$. Then $x_1 \le 0$ and $x_n \ge 0$.

Let $f(x) := \frac{1}{\frac12x + \frac12\sqrt{x^2 + 4} + n - 1}$. We have $$f''(x) = \frac{2(x + \sqrt{x^2 + 4})^3 - 16(n - 1)}{(x + \sqrt{x^2 + 4} + 2n - 2)^3 (x^2 + 4)^{3/2}}.$$ Let $q := \sqrt[3]{n-1} - \frac{1}{\sqrt[3]{n-1}}$. We have, $f''(q) = 0$, and $f''(x) < 0$ on $(-\infty, q)$, and $f''(x) > 0$ on $(q, \infty)$. Thus, $f(x)$ is concave on $(-\infty, q]$, and convex on $[q, \infty)$.

We split into two cases.

Case 1. $x_n \le q$

By Jensen's inequality, we have $$\sum_{i=1}^n f(x_i) \le n f\left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right) = nf(0) = 1.$$

Case 2. $x_n > q$

There exists $1 \le k \le n-1$ such that $x_k < q \le x_{k+1}$.

By Karamata's inequality,, since $(x_{k+1}, \cdots, x_n)$ is majorized by $(q, \cdots, q, x_{k+1}+\cdots + x_n - (n-k-1)q)$, we have $$\sum_{i=k+1}^n f(x_i) \le (n - k - 1)f(q) + f\Big(x_{k+1} + \cdots + x_n - (n-k-1)q\Big). \tag{2}$$

From (1) and (2), it suffices to prove that $$\sum_{i=1}^k f(x_i) + (n - k - 1)f(q) + f\Big(x_{k+1} + \cdots + x_n - (n-k-1)q\Big) \le 1. \tag{3}$$

By Jensen's inequality, we have $$\sum_{i=1}^k f(x_i) + (n - k - 1)f(q) \le (n-1)f\left(\frac{x_1 + \cdots + x_k + (n-k-1)q}{n-1}\right). \tag{4}$$

From (3) and (4), it suffices to prove that $$(n-1)f\left(\frac{-u}{n-1}\right) + f(u) \le 1 \tag{5}$$ where $u := x_{k+1} + \cdots + x_n - (n-k-1)q$.

Let $w := \frac12 u + \frac12\sqrt{u^2 + 4}$. Then $w > 0$ and $u = w - \frac{1}{w}$. (5) is written as $$(n-1)f\left(\frac{-w + 1/w}{n-1}\right) + \frac{1}{w + n - 1} \le 1$$ which is equivalent to \begin{align*} \frac12\sqrt{\frac{(w - 1/w)^2}{(n-1)^2} + 4} \ge \frac{w - 1/w}{2(n-1)} - (n - 1) + \frac{n - 1}{1 - \frac{1}{w + n - 1}}. \end{align*} It suffices to prove that \begin{align*} \left(\frac12\sqrt{\frac{(w - 1/w)^2}{(n-1)^2} + 4}\right)^2 \ge \left(\frac{w - 1/w}{2(n-1)} - (n - 1) + \frac{n - 1}{1 - \frac{1}{w + n - 1}}\right)^2 \end{align*} or $$\frac{(w - 1)^2(n-2)}{w (w + n - 2)^2} \ge 0$$ which is true.

We are done.

Answer 3

Counter example $\{a_{n}=1, for\ all\ natural\ n\}$.

Then each finite sum is equal n on both sides of the presumption: $\sum_{i=1}^{n}a_{i}=\sum_{i=1}^{n}\frac{1}{a_{i}}=\frac{n(n+1)}{2}=\{1,3,6,...\}$ then $\sum_{i=1}^{n}\frac{1}{a_{i}+n-1}=\sum_{i=1}^{n}\frac{1}{n}={1,\frac{3}{2},\frac{11}{6},...}=HarmonicNumber(k)$ that is from the second term on not equal.

This is a true counterexample since there is restriction other than the identity posed.