Given positive numbers $x_1,…,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$

Question

Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.

I tried solving this question through substitution, e.g. $a_1=\sqrt[n]{x_1},...,a_n=\sqrt[n]{x_n}$, hence $a_1...a_n=1$. Also for $k=1,2,...,n$ we have that $x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}$.

This is where I got stuck. My intuition tells me that I should be able to finish it off from here using AM-GM, however it isn't working out for me. Could you please explain to me how to finish the question off?

Answer 1

One can use Lagrange multipliers for: $$L = \sum_{k=1}^{n} 1/(n-1 + x_{k}) - \lambda(\prod_{k=1}^{n}x_{k} - 1)$$ One has: $$\frac{dL}{dx_{k}} = -\frac{x_{k}}{(n-1+ x_{k})^{2}} -\lambda\prod_{j\neq k}x_{j}$$ And when this is zero, one can multiply both sides by $x_{k}$ to get: $$\frac{x_{k}^{2}}{(n-1+ x_{k})^{2}} =\lambda\prod_{k}x_{k}$$ Because of condition on product of $x_{k}'s$ all $x_{k}$ must be the same and equal to 1 at the maximum.

Answer 2

According to page 8, Problem 25 in [1], the equivalent problem was published by Vasile Cirtoaje on Gazeta Matematica, Seria B, No. 10, 1991.

Five solutions are given in [1]. Here is one of them:

Let $r = \frac{n-1}{n}$. By AM-GM, we have, for $i=1, 2, \cdots, n$, $$\sum_{j\ne i} x_j^r \ge (n-1) \left(\prod_{j\ne i} x_j\right)^{r/(n-1)} = (n-1) x_i^{r - 1}$$ which results in $x_1^r + x_2^r + \cdots + x_n^r \ge (n-1)x_i^{r - 1} + x_i^r$ and $$\frac{x_i}{x_i + n-1} \ge \frac{x_i^r}{x_1^r + x_2^r + \cdots + x_n^r}. \tag{1}$$ Summing up (1) for $i = 1, 2, \cdots, n$, we have $$\sum_{i=1}^n \frac{x_i}{x_i + n-1} \ge 1$$ which is written as $$\sum_{i=1}^n \frac{1}{x_i + n-1} \le 1.$$ We are done.

Reference

[1] Vasile Cirtoaje, “Algebraic Inequalities: Old and New Methods”, 2006.

Answer 3

EDIT: As pointed out, this answer contains a flawed application of Cauchy-Schwarz. In particular, the L.H.S. of the inequality can be zero, while the R.H.S. is positive, an absurdity. I tried to remedy this flaw but ultimately failed. Hence, I have come up with a new approach that proceeds by a "smoothing principle" argument. Any feedback on the new proof will be much appreciated.

Note: In your post, you claimed the following:

$x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}.$

I do not see how the last equality follows. If it were true, then we would have $a_k=\dfrac{1}{a_1a_2...a_n}=1$, which is not necessarily the case. But anyways, it is possible to solve the problem using AM-GM and Cauchy-Schwarz.

Firstly, the case where $x_1=x_2=...=x_n=1$ is trivial. We only consider the case where $\exists \ i \in \{1,2,..,n\}$ such that $x_i \neq 1$. Note that:

$$\dfrac{1}{n-1+x_1} + \dfrac{1}{n-1+x_2} + ... + \dfrac{1}{n-1+x_n} \leq 1$$ \begin{align} & \iff \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_1}\right) + \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_2}\right) + ... + \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_n}\right) \geq 0 \\ & \iff \dfrac{x_1-1}{n(n-1+x_1)} + \dfrac{x_2-1}{n(n-1+x_2)} + ... + \dfrac{x_n-1}{n(n-1+x_n)} \geq 0 \\ & \iff \dfrac{x_1-1}{n-1+x_1} + \dfrac{x_2-1}{n-1+x_2} + ... + \dfrac{x_n-1}{n-1+x_n} \geq 0 \ (\because \dfrac{1}{n} > 0 ) \\ \end{align} By Cauchy-Schwarz, $$\left(\dfrac{x_1-1}{n-1+x_1} + \dfrac{x_2-1}{n-1+x_2} + ... + \dfrac{x_n-1}{n-1+x_n} \right) \left[(x_1-1)(n-1+x_1) + ... + (x_n-1)(n-1+x_n) \right]$$ $$\geq \left[(x_1-1)+(x_2-1) + ... + (x_n-1) \right]^2$$ The R.H.S. of the preceding inequality is clearly non-negative. It thus suffices to prove that: \begin{align} & (x_1-1)(n-1+x_1) + (x_2-1)(n-1+x_2) + ... + (x_n-1)(n-1+x_n) > 0 \\ & \iff n(x_1-1)+(x_1-1)^2 + n(x_2-1)+(x_2-1)^2 + ... + \ n(x_n-1)+(x_n-1)^2 > 0 \\ & \iff (x_1-1)^2 + ... + (x_n-1)^2 + n\left[(x_1-1) + ... + (x_n-1) \right] > 0. \\ \end{align}

Since $\exists \ i \in \{1,2,..,n\}$ such that $x_i \neq 1$, it is obvious that $(x_1-1)^2 + ... + (x_n-1)^2 > 0$. Now, we claim that $n\left[(x_1-1) + ... + (x_n-1) \right] \geq 0 $, which concludes our proof. To see why this is true, note that:

\begin{align} n\left[(x_1-1) + ... + (x_n-1) \right] & = n(x_1 + ... + x_n -n) \\ & \geq n(n \sqrt[^n]{x_1...x_n} - n) \ (\text{By applying AM-GM inequality})\\ & = n(n-n) \ (\because x_1...x_n=1) \\ &=0. \end{align}

Answer 4

The case $x_1=x_2=...=x_n=1$ is trivial, and for $n >2$, is the only case when equality holds. We prove this below.

Suppose the $x_i$s are not all $1$. Then, there exists $i,j \in \{1,2,..,n\}, i \neq j$, such that $x_i<1<x_j$. Replace the pair $(x_i \ ,x_j)$ with $(x_i'\ ,x_j')$ , such that:

$$x_i'=1, \ x_j'=x_ix_j.$$ $x_i'$ and $x_j'$ have the same product as $x_i$ and $x_j$. Consider the sum

$$\dfrac{1}{n-1+x_i} + \dfrac{1}{n-1+x_j} = \dfrac{x_i+x_j+2(n-1)}{(n-1)^2 + (n-1)(x_i+x_j) + x_ix_j}.$$

It would be ideal if the above sum was strictly smaller than $\dfrac{1}{n-1+x_i'} + \dfrac{1}{n-1+x_j'} = \dfrac{1}{n} + \dfrac{1}{n-1+x_ix_j}.$ (By our replacement, we increase the number of $x_i$s which are equal to $1$, while strictly increasing the L.H.S. of the given inequality and preserving the given constraint. Eventually, all the $x_i$s will equal $1$, and the inequality becomes an equality.) Attempting to proceed along this line,

\begin{align} & \dfrac{1}{n-1+x_i} + \dfrac{1}{n-1+x_j} < \dfrac{1}{n} + \dfrac{1}{n-1+x_ix_j} \\ & \iff \dfrac{1}{n-1+x_i} - \dfrac{1}{n} < \dfrac{1}{n-1+x_ix_j} - \dfrac{1}{n-1+x_j} \\ & \iff \dfrac{1-x_i}{n(n-1+x_i)} < \dfrac{x_j-x_ix_j}{(n-1+x_ix_j)(n-1+x_j)} \\ & \iff \dfrac{1}{n(n-1+x_i)} < \dfrac{x_j}{(n-1+x_ix_j)(n-1+x_j)} \\ & \iff (n-1+x_ix_j)(n-1+x_j) < nx_j(n-1+x_i) \\ & \iff x_ix_j^2 - x_ix_j + nx_j - x_j + n^2-2n+1 - n^2x_j + nx_j < 0 \\ & \iff n^2(1-x_j) - x_ix_j(1-x_j) - 2n(1-x_j) + (1-x_j) < 0 \\ & \iff n^2-x_ix_j -2n + 1 >0 \\ & \iff x_ix_j < (n-1)^2. \end{align} Unfortunately, it is possible to have $x_ix_j \geq (n-1)^2$. Fortunately, this issue can be circumvented. W.L.O.G. let $x_1 \leq x_2 ... \leq x_n , x_1 < 1 < x_n$. There are $2$ cases:

Case $1$: $x_1x_n \geq (n-1)^2 \Rightarrow \dfrac{x_1+x_n+2(n-1)}{(n-1)^2 + (n-1)(x_1+x_n) + x_1x_n } \leq \dfrac{x_1+x_n+2(n-1)}{2(n-1)^2 + (n-1)(x_1+x_n)} = \dfrac{1}{n-1}.$ Since $\dfrac{1}{n-1+x_n}$ is positive, we conclude that $\dfrac{1}{n-1+x_1} < \dfrac{1}{n-1}$. But the largest term in the sum $ \displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i}$ is $\dfrac{1}{n-1+x_1}$. Thus, $$\displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i} \leq \dfrac{1}{n-1} + \displaystyle \sum_{i=2}^{n-1} \dfrac{1}{n-1+x_i} < \dfrac{1}{n-1} + \dfrac{n-2}{n-1} =1. $$

Case $2$: $x_1x_n < (n-1)^2 \Rightarrow \dfrac{1}{n-1+x_1} + \dfrac{1}{n-1+x_n} < \dfrac{1}{n} + \dfrac{1}{n-1+x_1x_n}$. Letting $x_1'=1, x_n'=x_1x_n$ and $x_j'=x_j \ \forall \ j \notin \{1,n\},$ we have $\displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i} < \displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i'}.$ And we are done by a repeated application of this "smoothing" procedure! (At each stage, W.L.O.G. re-order the variables $x_i$s from smallest to largest, since the previous round of "smoothing" disrupted the original order.) If, at any stage, the product of the smallest term with the largest term is greater than or equal to $(n-1)^2$, we are done by Case $1$. Otherwise, we continually increase the L.H.S. and eventually end up with all the $x_is$ equal to $1$, giving the equality case L.H.S. $=1$.

Answer 5

Partial answer :

Remarking that the function $f(x)=\frac{e^x}{e^x(n-1)+1}$ is concave on $(-\ln(n-1),\infty)$ and rewriting the inequality like :

$$S=\sum_{i=1}^{n}\frac{e^{x_i}}{e^{x_i}(n-1)+1}\leq 1$$

We apply Jensen's inequality to get the value $1$ since $e^{\sum_{i=1}^{n}x_i}=1$ and $x_i\in(-\ln(n-1),\infty)$