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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?

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    $\begingroup$ It was mentioned in the comments under the answer. But I will also add a direct link to the search in Approach Zero. I do not see something close to this among the top results in SearchOnMath. Both of them are mentioned in the faq post: How to search on this site? $\endgroup$ – Martin Sleziak Jan 1 at 10:50
  • $\begingroup$ Precisely this one is a showcase in Thomas Mildorf's "Olympiad inequalities" (2005): It is no. 21, tagged "Romania 1999", and solved by the $n-1$ equal variables method. $\endgroup$ – Hanno Feb 6 at 8:40
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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.

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  • $\begingroup$ Please check the term $-\frac{x_{k}}{(n-1+ x_{k})^{2}}$ in the derivative expression. $\endgroup$ – River Li Jan 8 at 2:22
  • $\begingroup$ I think it should be $\frac{dL}{dx_{k}} = -\frac{1}{(n-1+ x_{k})^{2}} -\lambda\prod_{j\neq k}x_{j}$ which results in $\frac{x_{k}}{(n-1+ x_{k})^{2}} = -\lambda\prod_{k}x_{k} = -\lambda$. However, from $\frac{x_{k}}{(n-1+ x_{k})^{2}} = -\lambda$ and $\prod_k x_k = 1$, we can not get $x_1 = x_2 = \cdots = x_n = 1$ (with $\lambda = -\frac{1}{n^2}$). For example, $n=3$, $x_1 = x_2 = \frac{1}{4}, x_3 = 16$ and $\lambda = -\frac{4}{81}$ also satisfies $\frac{x_{k}}{(n-1+ x_{k})^{2}} = -\lambda$ and $\prod_k x_k = 1$. $\endgroup$ – River Li Jan 14 at 1:28
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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.

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  • $\begingroup$ I am surprised why you didnt post vasc's third solution using C-S,i felt it to be most elegant than this one...... $\endgroup$ – Albus Dumbledore Jan 13 at 8:29
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    $\begingroup$ @AlbusDumbledore $\frac{x_i}{x_i + n-1} \ge \frac{x_i^r}{x_1^r + x_2^r + \cdots + x_n^r}$ is the so-called isolated fudging, see yufeizhao.com/olympiad/wc08/ineq.pdf We may use calculus to determine the value of $r = \frac{n-1}{n}$. $\endgroup$ – River Li Jan 13 at 8:53
  • $\begingroup$ Riverli thanks i didnt know that method .............But i still felt vasc's third solution was better than the others..... $\endgroup$ – Albus Dumbledore Jan 13 at 8:57
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    $\begingroup$ @SeeHai You should first make it homogeneous, that is, $f(x_1,...,x_n)=\dfrac{x_1}{x_1+(n-1)\sqrt[n]{\prod_i x_i}} - \dfrac{x_1^r}{x_1^r + ... + x_n^r}$. $\endgroup$ – River Li Jan 14 at 17:02
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    $\begingroup$ @SeeHai Since there is a constraint $\prod_i x_i = 1$, after homogenization, we do not need to consider the constraint. To determine $r$ is the first step. The 2nd step is to check if $\dfrac{x_1}{x_1+n-1} - \dfrac{x_1^r}{x_1^r + ... + x_n^r}\ge 0$ for this $r$. Sometimes, such a $r$ does not work. $\endgroup$ – River Li Jan 15 at 9:37
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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}

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  • $\begingroup$ Please check the expression after "By Cauchy-Schwarz": for example, $n=2$, $x_1 = 2, x_2 = 1/2$. $\endgroup$ – River Li Jan 11 at 2:03
  • $\begingroup$ Yes, you are right! I would have to rethink how to fix this. $\endgroup$ – See Hai Jan 11 at 5:05
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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. $=1$.

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  • $\begingroup$ It is a nice solution. (+1) $\endgroup$ – River Li Jan 13 at 2:03
  • $\begingroup$ @RiverLi thank you too for reading through both my solutions and providing feedback! $\endgroup$ – See Hai Jan 13 at 3:19
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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 $x_i\in(-\ln(n-1),\infty)$

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  • $\begingroup$ How do you apply Jensen? $\endgroup$ – River Li Jan 14 at 1:03
  • $\begingroup$ @RiverLi think order !(I want to preserve the hint so sorry if it's enigmatic) $\endgroup$ – Erik Satie Jan 14 at 18:23
  • $\begingroup$ I think we can not directly apply Jensen here. Show us if you can prove the inequality by simply applying Jensen. $\endgroup$ – River Li Jan 14 at 18:33
  • $\begingroup$ @RiverLi Can you underline more precisely what is not right ? $\endgroup$ – Erik Satie Jan 15 at 11:52
  • $\begingroup$ Would you please write more steps for "by a judicious choice apply Jensen's inequality to n−1 terms"? $\endgroup$ – River Li Jan 15 at 12:01

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