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

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$$$$\tag2\frac{1}{a_{1}+n-1}+\frac{1}{a_{2}+n-1}+\dots+\frac{1}{a_{n}+n-1}<1$$$$

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 $$$$\sum_{i=1}^n\frac{1}{a_{i}+n-1}+\sum_{i=1}^n\frac{1}{\frac{1}{a_{i}}+n-1}<2$$$$ If $$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)?

• I posted a faulty answer earlier - but this definitely yields to Lagrange multiplier which implies that the only local maximum is at $a_1=a_2=...=a_n.$ The only obstacle would be to show that the inequality holds on the "boundaries" of first quandrant of $\mathbb{R}^n.$ Commented Dec 30, 2023 at 20:32
• to get rid of the "boundary", you can take $u_i=a_i-a_i^{-1}$ as the variable, which does not have any positivity constraints, only the $\sum_{i=1}^n u_i = 0$. Commented Dec 30, 2023 at 21:18

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

$$$$L(b_i, \lambda) = \sum_{i=1}^n \frac{1}{(n-1) + \frac{b_i + \sqrt{b_i^2 + 4}}{2}} - \lambda \sum_{i=1}^n b_i, \qquad \sum_{i=0}^n b_i = 0$$$$

$$$$\lambda = -\frac{1}{\sqrt{b_i^2 + 4}} \frac{\frac{b_i + \sqrt{b_i^2 + 4}}{2}}{\left((n-1)+\frac{b_i + \sqrt{b_i^2 + 4}}{2} \right)^2}$$$$

Without solving one can see that $$b_i(\lambda)=b_j(\lambda), \forall i,j$$. Implicitly solving and plugging into the constraint

$$$$0 = \sum_{i=1}^n b_i = \sum_{i=1}^n b_i(\lambda) = n \cdot b(\lambda)$$$$

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 $$f: \mathbb{R}^{n-1} \to \mathbb{R}$$

$$$$(b_1, \dots, b_{n-1}) \mapsto \frac{1}{(n-1) + \frac{-(b_1 + \dots + b_{n-1}) + \sqrt{(b_1 + \dots + b_{n-1})^2 + 4}}{2}} + \sum_{j=1}^{n-1}\frac{1}{(n-1) + \frac{b_j + \sqrt{b_j^2 + 4}}{2}}$$$$

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.

• Thanks, I need to adjust the statement, I added n>1 and changed $\leqslant$ to $<$. This will rule out $a_i=1,\forall i$ Commented Dec 31, 2023 at 2:34
• "Without solving one can see that $b_i(\lambda)=b_j(\lambda), \forall i,j$" How to see that the equation in $b_i$ has unique solution? Commented Dec 31, 2023 at 2:55
• You still need $\leq 1$, since e.g. for $n=2$ if $a+b = 1/a + 1/b$ it follows that $a= -b$ or $a = 1/b$. As $a,b>0$ it follows that necessarily $a=1/b$ but then $1/(a+1) + 1/(b+1) = (b+1)/(b+1) = 1$ Commented Dec 31, 2023 at 4:07
• Thanks. I added $n>2$. This will make $<1$ possible in (2). Commented Dec 31, 2023 at 4:38
• In "Without solving one can see that $b_i(\lambda)=b_j(\lambda), \forall i,j$" Why is the map $b_i\mapsto\frac{1}{\sqrt{b_i^2 + 4}} \frac{\frac{b_i + \sqrt{b_i^2 + 4}}{2}}{\left((n-1)+\frac{b_i + \sqrt{b_i^2 + 4}}{2} \right)^2}$ injective on $\Bbb R$? Commented Dec 31, 2023 at 4:43

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.

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.

• I do not understand why this was downvoted. Commented Dec 31, 2023 at 6:04
• downvote is probably unwarranted but this is missing the meat of the question. OP phrased it poorly but it's obvious that the only interesting question is whether $(1)$ implies $(2).$ Also, your example does not work because you have to fix $n$ first and second sum is also always $1.$ Commented Dec 31, 2023 at 16:27