Here's the solution from my recent handout on the Method of Balancing Coefficients. Taking $x_1=x_2=...=x_n=1$, we find that $k_n \geq \frac{(2n-1)}{(n-1)^2}$. We will prove that $k_n = \frac{(2n-1)}{(n-1)^2}$. So it remains to show that $$\sum_{i=1}^{n}\frac{1}{\sqrt{1+ \frac{(2n-1)x_k}{(n-1)^2}}}\leq n-1$$Let $x_1x_2....x_n=1$. Suppose this inequality doesn't hold for a certain system of $n$ numbers with product $1$ and $x_1,x_2,..,x_n >0$. Thus we can find a number $M >n-1$ and some numbers $a_i >0$ which add up to $1$, such that $$\frac{1}{\sqrt{1+ \frac{(2n-1)x_k}{(n-1)^2}}}= Ma_k$$Thus $a_k < \frac{1}{n-1}$ and we have $$1=\prod_{i=1}^{n}\left[\frac{(n-1)^2}{2n-1}\left(\frac{1}{M^2a_k^2}-1\right)\right]$$which implies $$\left(\frac{(2n-1)}{(n-1)^2}\right)^n<\prod_{i=1}^{n}\left(\frac{1}{(n-1)^2a_k^2}-1\right).$$Now, denote $1-(n-1)a_k=b_k >0$ and observe that $\displaystyle \sum_{i=1}^{n}b_k=1$. Also, the above inequality becomes $$(n-1)^{2n}\prod_{i=1}^{n}b_k\prod_{i=1}^{n}(2-b_k)>(2n-1)^n\left(\prod_{i=1}^{n}(1-b_k)\right)^2$$Because from AM-GM we have $$\prod_{i=1}^{n}(2-b_k) \leq \left(\frac{(2n-1)}{n}\right)^n$$our assumption leads to $$\left(\prod_{i=1}^{n}(1-b_k)\right)^2 < \frac{(n-1)^{2n}}{n^n}b_1b_2 \ldots b_n$$so, it is enough to prove that for any positive numbers $a_1,a_2,\ldots,a_n$ the inequality $$\prod_{i=1}^{n}(a_1+a_2+\ldots+a_{k-1}+a_{k+1}+\ldots+a_n)^2 \geq \frac{(n-1)^{2n}}{n^n}a_1a_2 \ldots a_n(a_1+a_2+\ldots +a_n)^n$$holds. This stronger inequality will be proved by induction. For $n=3$ it follows from the fact $$\frac{\left(\prod(a+b)\right)^2}{abc} \geq \frac{\left(\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\right)^2}{abc} \geq \frac{64}{27}(a+b+c)^3$$Now, suppose the inequality is true for all systems of $n$ numbers. Let $a_1,a_2,...,a_{n+1}$ be positive real numbers. And as the inequality is homogeneous and symmetric , we may assume that $a_1 \leq a_2 \leq a_3 \leq \ldots \leq a_{n+1}$ and also that $a_1+a_2+...+a_{n+1}=1$. Applying the inductive hypothesis we get the inequality $$\prod_{i=1}^{n}(1-a_i)^2 \geq \frac{(n-1)^{2n}}{n^n}a_1a_2 \ldots a_n$$To prove the inductive step we must prove that $$\prod_{i=1}^{n}(a_{n+1}+1-a_i)^2 \geq \frac{n^{2n+2}}{(n+1)^{(n+1)}}a_1a_2\ldots a_{n+1}(1+a_{n+1})^{n+1}$$thus it is enough to prove the stronger inequality $$\prod_{i=1}^{n}\left(1+\frac{a_{n+1}}{(1-a_i)}\right)^2 \geq \frac{n^{3n+2}}{(n-1)^{2n}(n+1)^{n+1}}a_{n+1}(1+a_{n+1})^{n+1}$$Now, using Huygens inequality and AM-GM, we find that $$\prod_{i=1}^{n}\left(1+\frac{a_{n+1}}{(1-a_i)}\right)^2 \geq \left(1+\frac{a_{n+1}}{\sqrt{\prod_{i=1}^{n}(1-a_i)}}\right)^{2n} \geq \left(1+ \frac{na_{n+1}}{n-1}\right)^{2n}$$and so we are left with the inequality $$\left(1+ \frac{na_{n+1}}{n-1}\right)^{2n}\geq \frac{n^{3n+2}}{(n-1)^{2n}(n+1)^{n+1}}a_{n+1}(1+a_{n+1})^{n+1}.$$If $a_{n+1} \geq max(a_1,a_2,..,a_n) \geq \frac{1}{n}$. So we can put $\dfrac{n(1+a_{n+1})}{(n+1)}= 1+x$, where $x$ is nonnegative. So, the inequality becomes $$\left(1+\frac{x}{n(x+1)}\right)^{2n} \geq \frac{1+(n+1)x}{(x+1)^{n-1}}$$Using Bernoulli inequality, we find immediately that $$\left(1+\frac{x}{n(x+1)}\right)^{2n} \geq \frac{3x+1}{x+1}$$Also, $(1+x)^{n-1} \geq 1 +(n-1)x$ and so it is enough to prove that $$\frac{3x+1}{x+1} \geq \frac{1+(n+1)x}{1+(n-1)x}$$which is trivial. So, we have reached a contradiction assuming the inequality doesn't hold for a certain system of $n$ number with product $1$, which shows our inequality is true for $k_n = \dfrac{(2n-1)}{(n-1)^2}$.

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