first of all I would like to point out that i am able to solve this inequality using standard techniques such as Rearrangement inequality , C-S , Jensen's....
However I'm in the middle of learning a new technique in inequality solving , which is the smoothing technique and I am having trouble fully understanding it.
Here is an example taken from this PDF File :https://artofproblemsolving.com/articles/files/KedlayaInequalities.pdf
Theorem 1 (AM-GM). Let $x_1, \cdots, x_n $ be positive real numbers. Then
$$\dfrac{x_1+\cdots+x_n}{n} \geq \sqrt[n]{x_1\cdots x_n},$$
with equality if and only if $x_1=\cdots=x_n.$
Proof. We will make a series of substitutions that preserve the left-hand side while strictly increasing the right-hand side. At the end, the $x_i$ will all be equal and the left-hand side will equal the right-hand side; the desired inequality will follow at once. (Make sure that you understand this reasoning before proceeding!)
If the $x_i$ are not already all equal to their arithmetic mean, which we call $a$ for convenience, then there must exist two indices, say $i$ and $j$, such that $x_i<a<x_j$. (If the $x_i$ were all bigger than $a$, then so would be their arithmetic mean, which is impossible; similarly if they were all smaller than $a$.) We will replace the pair $x_i$ and $x_j$ by
$$x_i'=a, x_j'=x_i+x_j-a;$$
by design, $x_i'$and $x_j'$have the same sum as $x_i$ and $x_j$, but since they are close together, their product is larger. To be precise,
$$a(_i+x_j-a)=x_ix_j+(x_j-a)(a-x_i)>x_ix_j$$
because $x_j-a$ and $a-x_i$ are positive numbers.
By this replacement, we increase the number of the $x_i$ which are equal to $a$, preserving the left-hand side of the desired inequality by increasing the right-hand side. As noted initially, eventually this process ends when all of the $x_i$ are equal to $a$, and the inequality becomes equality in that case. It follows that in all other cases, the inequality holds strictly. $\square$
Now Here is my attempt to solve our problem:
$$\sum_{i=1}^{n}\frac{x_{i}}{\sqrt{1-x_{i}}} \geq \sqrt{\frac{n}{n-1}}$$ subject to the constraint $\sum_\limits{i=1}^{n} x_{i} = 1 $
First of all , let $a=\frac{x_{1}+x_{2}+x_{3} \cdots +x_{n}}{n}$ (The Arithmetic mean of $x_{i}$) and $L=\sum_\limits{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}}$ There there exists two indices i and j , st $x_{i}<a<x_{j}$ (Proof shown in the Image above )
Now we will replace the pair $x_{i}$ and $x_{j}$ by $x'_{i}=a,x'_{j}=x_{i}+x_{j}-a$
In Doing so , we get : $$\frac{x_{i}}{1-x_{i}} \leq \frac{a}{\sqrt{1-a}}$$
Which means $$L\leq n\frac{a}{\sqrt{1-a}}$$ But $a=\frac{1}{n}$ So:$$L \leq \sqrt{\frac{n}{n-1}}$$
Which contradicts the Problem .
I'd like to know where did i fault , and possibly get some lead in the right way using the smoothing technique correctly.
Any communicated references or articles about the smoothing technique would be greatly appreciated
and if one manages to provide an answer with the smoothing technique , please submit.
Thanks to the community , and have a good day everyone!
