# Inequality with smoothing technique:$\sum_{i=1}^{n}\frac{x_{i}}{\sqrt{1-x_{i}}}\geq \sqrt{\frac{n}{n-1}}$ subject to $\sum_{i=1}^{n} x_{i}=1$

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 $$x_i. (If the $$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} (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:$$L \leq \sqrt{\frac{n}{n-1}}$$