$$LHS=\sum\limits_{cyc-i}{\dfrac{a_i}{\sqrt{1-a_i}}}=\sum_{cyc-i}{\dfrac{a_i^2}{a_i\sqrt{1-a_i}}}\geq \dfrac{1}{\sum\limits_{cyc}{a_i\sqrt{1-a_i}}}\geq \dfrac{1}{\left(a_1+a_2+\cdots+a_n\right)\sqrt{\dfrac{\sum\limits_{cyc-i}{a_i-a_i^2}}{a_1+a_2+\cdots+a_n}}}$$$$\geq \dfrac{1}{\sqrt{1-\dfrac{\left(a_1+a_2+\cdots+a_n\right)^2}{n}}}=\sqrt{\dfrac{n}{n-1}}\geq RHS$$last one is trivial by Cauchy One cannot generalize this problem via $a_1+a_2+\cdots+a_n=\lambda$ and then putting $\lambda$ to $1$, cuz substituting $a_1=\lambda b_1$ transforms it to the original one with the same information.

ehuseyinyigit wrote: $$LHS=\sum\limits_{cyc-i}{\dfrac{a_i}{\sqrt{1-a_i}}}=\sum_{cyc-i}{\dfrac{a_i^2}{a_i\sqrt{1-a_i}}}\geq \dfrac{1}{\sum\limits_{cyc}{a_i\sqrt{1-a_i}}}\geq \dfrac{1}{\left(a_1+a_2+\cdots+a_n\right)\sqrt{\dfrac{\sum\limits_{cyc-i}{a_i-a_i^2}}{a_1+a_2+\cdots+a_n}}}$$$$\geq \dfrac{1}{\sqrt{1-\dfrac{\left(a_1+a_2+\cdots+a_n\right)^2}{n}}}=\sqrt{\dfrac{n}{n-1}}\geq RHS$$last one is trivial by Cauchy One cannot generalize this problem via $a_1+a_2+\cdots+a_n=\lambda$ and then putting $\lambda$ to $1$, cuz substituting $a_1=\lambda b_1$ transforms it to the original one with the same information. what did u do in the fourth statement sry I'm confused

Here $$\sum_{cyc-i}{\dfrac{a_i^2}{a_i\sqrt{1-a_i}}}\geq \dfrac{1}{\sum\limits_{cyc}{a_i\sqrt{1-a_i}}}$$I used Bersgtröm's Inequality (Titu's lemma). And here $$\dfrac{1}{\sum\limits_{cyc}{a_i\sqrt{1-a_i}}}\geq \dfrac{1}{\left(a_1+a_2+\cdots+a_n\right)\sqrt{\dfrac{\sum\limits_{cyc-i}{a_i-a_i^2}}{a_1+a_2+\cdots+a_n}}}$$there is an application of Jensen's Inequality.

By Jensen's Inequality we have $$\sum\limits_{cyc-i}{\dfrac{a_i}{\sqrt{1-a_i}}}\geq \sqrt{\dfrac{n}{n-1}}$$By Cauchy's Inequality we have $$\sqrt{\dfrac{n}{n-1}}\geq RHS$$Done.

Same as #6.

Let $$a_1 \geq a_2 \geq \cdots \geq a_n$$Then $$\frac{1}{\sqrt{1-a_1}} \geq \frac{1}{\sqrt{1-a_2}} \geq \cdots \geq \frac{1}{\sqrt{1-a_n}}$$By Chebyshev, $$\sum_{i=1}^n \frac{a_i}{\sqrt{1-a_i}} \geq \frac{1}{n} \left (\sum_{i=1}^n a_i \right) \left (\sum_{i=1}^n \frac{1}{\sqrt{1-a_i}} \right)$$By Titu's lemma $$\sum_{i=1}^n \frac{1}{\sqrt{1-a_i}} \geq \frac{n^2}{\sum_{i=1}^n \sqrt{1-a_i}}$$So $$\frac{1}{n} \left (\sum_{i=1}^n a_i \right) \left (\sum_{i=1}^n \frac{1}{\sqrt{1-a_i}} \right) \geq \frac{1}{n} \left (\sum_{i=1}^n a_i \right) \left(\frac{n^2}{\sum_{i=1}^n \sqrt{1-a_i}} \right) = \left (\sum_{i=1}^n a_i \right) \left(\frac{n}{\sum_{i=1}^n \sqrt{1-a_i}} \right) = \frac{n}{\sum_{i=1}^n \sqrt{1-a_i}} $$Now by Cauchy $$\frac{n}{\sum_{i=1}^n \sqrt{1-a_i}} \geq \frac{n}{\sqrt{n} \sqrt{n-(a_1 + \cdots + a_n)}} = \frac{\sqrt{n}}{\sqrt{n-1}} = \frac{\sqrt{n}\sqrt{a_1 + \cdots + a_n}}{\sqrt{n-1}} \geq \frac{\sum_{i=1}^n \sqrt{a_i}}{\sqrt{n-1}} $$and we're done? Is this correct?