Translation in a normal language: Let $n\ge 3$ be an integer. Given that $\sum_{i=1}^n \frac{1}{x_i+n-1} =1,$ where $x_i$'s are positive reals, find the minimum value of $$\sum_{i=1}^n\left( \frac{1}{x_i} -x_i \right)$$

The following approach works, thou some calculations are needed. Let $$f(x_1,x_2,\dots,x_n):=\sum_{i=1}^n\left( \frac{1}{x_i} -x_i \right)$$$f(1,1,\dots,1)=0$, so the min value (if exists) must be at most $0$. First see that when some $x_i$ is large enough, $f$ becomes also large. We consider the set $D:= \{(x_1,x_2,\dots,x_n): 0<x_i\leq A\}, i=1,2,\dots,n\}$, where $A>0$ is large enough. It's a bounded set. Note that $f(x_1,\dots,x_n)$ is a convex function, that is more than $1$ in the boundary of $D$, which means it's equivalent to search for min of $f$ on some compact subset $K$ of $D$. We denote $F=\{(x_1,x_2,\dots,x_n): P(x_1,x_2,\dots,x_n)=1\}\}$, where $P(\dots)$ is as in the post N1. Let $K'=K\cap F$. Since $F$ is closed, $K'$ is also compact. (it is not empty, since $(1,1,\dots,1)\in K'$). Now, $f$ over $K'$ has a min value, which is attained at some point $(x^0_1,x^0_2,\dots,x^0_n)$. For the sake of contradiction, suppose not all $x^0_i$ are equal, WLOG let $x^0_1\neq x^0_2$. We vary only $x_1,x_2$, keeping $$ \frac{1}{x_1+n-1}+ \frac{1}{x_2+n-1}=\frac{1}{x^0_1+n-1}+ \frac{1}{x^0_2+n-1}=\mathrm{const}$$It remains to calculate that when $x_1=x_2$, the value of $f$ is less than $f(x^0_1,x^0_2,\dots)$. Contradiction. Hence, the min value of $f$ under the given restriction is $f(1,1,\dots1)=0$.

dgrozev wrote: It remains to calculate that when $x_1=x_2$, the value of $f$ is less than $f(x^0_1,x^0_2,\dots)$. Contradiction. Hence, the min value of $f$ under the given restriction is $f(1,1,\dots1)=0$. how do you do this?

Calculus. A routine calculation, thou not an elegant one.

Let $x_i \rightarrow \frac{(n-1)a_i}{S-a_i},$ where $a_i$'s are positive reals and $S=\sum_{i=1}^n a_i,$ the given inequality is equivalent to $$\sum_{i=1}^n{\frac{S-a_i}{a_i}} \ge (n-1)^2\sum_{i=1}^n{\frac{a_i}{S-a_i}} \Longleftrightarrow$$$$\sum_{i=1}^n\left(a_i\sum_{j=1 | j\neq i}^n{\frac{1}{a_j}}\right)\ge \sum_{i=1}^n\left(a_i\cdot \frac{(n-1)^2}{S-a_i}\right).$$The last inequality being true, as $\sum_{j=1 | j\neq i}^n{\frac{1}{a_j}}\ge \frac{(n-1)^2}{S-a_i}$ by CS in Engel's form.

First, we attempt to simplify the expression of $\dfrac{1}{x_i+n-1}$. Set $a_i$ equal to that expression; and so we get \[ x_i = \dfrac{1}{a_i}-n+1, \sum_{i=1}^{n} a_i = 1\]The inequality we're trying to prove becomes \[ \sum_{i=1}^{n} \dfrac{a_i}{\sum_{j=1}^{n}a_j - (n-1)a_i} \geq \sum_{i=1}^{n} \dfrac{\sum_{j=1}^{n}a_j}{a_i} + n-n^2 \]after getting homogenized. This looks a bit worse, but this form has an upside $-$ that we do not have a constraint on $a_i$. Now we substitute $b_i$ equal to the denominator on the $i^{th}$ term, and so solving for $a_i$ yields \[ \sum_{i=1}^{n} \dfrac{S-b_i}{(n-1)b_i} \geq \sum_{1\leq i,j \leq n}\dfrac{S-b_i}{S-b_j} - n(n-1) \]($S = \sum_{i=1}^{n} b_i$). After summing on the denominators on the $RHS$ (because it's what we do when seeing lots of equal denominators), we get \begin{align*} \sum_{i=1}^{n} \dfrac{S-b_i}{b_i} &\geq (n-1)\left( \sum_{i=1}^{n} \dfrac{(n-1)S}{S-b_i} - n(n-1) \right) \\ \sum_{1\leq i,j \leq n , i \ne j} \dfrac{b_i}{b_j} &\geq (n-1) \left( \dfrac{(n-1)b_i}{S-b_i} \right) \end{align*}Then it's easy to finish from here, by noting that

Solved with W. Olfram Let $y_i=\frac1{x_i+n-1}$. Then, $\sum_{i=1}^ny_i=1$ and $0<y_i<\frac1{n-1}$. Then, we have $$x_i+n-1=\frac1{y_i},$$so $$x_i=\frac1{y_i}+1-n$$and $$\frac1{x_i}=\frac{y_i}{1+y_i-ny_i}.$$Therefore, $$\frac1{x_i}-x_i=\frac{y_i}{1+y_i-ny_i}-\frac1{y_i}+n-1.$$Let $f(y)=\frac y{1+y-ny}-\frac1y+n-1$. Then, $$f''(y)=-\frac{2n-2}{(ny-y-1)^3}-\frac2{y^3}.$$This is equal to $0$ when we have \begin{align*} \frac{1-n}{(ny-y-1)^3}&=\frac1{y^3}\\ \sqrt[3]{1-n}&=\frac{y(n-1)-1}y\\ \sqrt[3]{1-n}&=n-1-\frac1y.\end{align*} This has exactly one solution between $0$ and $\frac1{n-1}$. Therefore, $f$ has one inflection point in $\left(0,\frac1{n-1}\right)$, so the minimum occurs when $n-1$ of the $y_i$ are equal by $n-1$ EV. Suppose that $y_1=y_2=\ldots=y_{n-1}$ and $y_n=1-(n-1)y_1$. Then, we have \begin{align*} \sum_{i=1}^n\left(\frac1{x_i}-x_i\right)&=\sum_{i=1}^n\left(\frac{y_i}{1+y_i-ny_i}-\frac1{y_i}+n-1\right)\\ &=(n-1)\left(\frac{y_1}{1+y_1-ny_1}-\frac1{y_1}\right)+\frac{1-(n-1)y_1}{1+(1-n)(1-(n-1)y_1)}-\frac1{1-(n-1)y_1}+n^2-n\\ &=\frac{(n-2)(n-1)(ny_1-1)^2}{y_1((n-1)^2y_1-n+2)}.\end{align*} This is negative only when $(n-1)^2y_1<n-2$. However, this means that $y_n=1-(n-1)y_1>1-\frac{n-2}{n-1}=\frac1{n-1}$, which is a contradiction. Therefore, $\sum_{i=1}^n\left(\frac1{x_i}-x_i\right)\geq0$, and equality holds when $x_1=x_2=\ldots=x_n=1$.