If $a_1,a_2,\dots ,a_n\ge 0$ such that $a_1+a_2+\dots a_n=1$ show that $\frac{a_1}{\sqrt{1-a_1}}+\frac{a_2}{\sqrt{1-a_2}}+\dots +\frac{a_n}{\sqrt{1-a_n}} \geq \frac{1}{\sqrt{n-1}}(\sqrt{a_1}+\sqrt{a_2}+\dots+ \sqrt{a_n})$.
Attempt
WLOG assume that $a_1\leq a_2\leq \dots \leq a_n$ it implies $\frac{1}{\sqrt{1-a_1}}\leq \frac{1}{\sqrt{1-a_2}}\leq \dots \leq \frac{1}{\sqrt{1-a_n}}$ now if we denote by $S$ the LHS of the inequality then by rearrangement inequality
$$S\geq \frac{a_2}{\sqrt{1-a_1}}+\frac{a_3}{\sqrt{1-a_2}}+\dots +\frac{a_1}{\sqrt{1-a_n}}$$
$$S \geq \frac{a_3}{\sqrt{1-a_1}}+\frac{a_4}{\sqrt{1-a_2}}+\dots +\frac{a_2}{\sqrt{1-a_n}}$$
$$\vdots$$
$$S \geq \frac{a_n}{\sqrt{1-a_1}}+\frac{a_{n-1}}{\sqrt{1-a_2}}+\dots +\frac{a_1}{\sqrt{1-a_1}}$$
it is $$(n-1)S\geq \left(\frac{1}{\sqrt{1-a_1}}+\frac{1}{\sqrt{1-a_2}}+\dots +\frac{1}{\sqrt{1-a_n}}\right) \left(a_1+a_2+\dots a_n \right)\geq$$
since $\frac{1}{1-a_i}\geq \frac{a_i}{1-a_i}\geq a_i$ we get $\frac{1}{\sqrt{1-a_i}}\geq \sqrt{\frac{a_i}{1-a_i}}\geq \sqrt{a_i}$
$$\geq (\sqrt{a_1}+\sqrt{a_2}+\dots \sqrt{a_n})(a_1+a_2+\dots a_n)$$
and since $a_i>\sqrt{a_i}$
$$(n-1)S\geq (\sqrt{a_1}+\sqrt{a_2}+\dots \sqrt{a_n})^2$$
But it not look like our inequality
any advice or help was useful thanks in advice.
Best Answer
Let $x \in (0,1)$, then notice that $f(x)=\frac{x}{\sqrt{1-x}} \implies f''(x)=\frac{4-x}{4(1-x)^{5/2}}>0.$ So by Jensen's inequality, we can write that $$S=\sum_{k=1}^n \frac{a_k}{\sqrt{1-a_k}}\ge n \frac{\sum_{k} a_k/n}{\sqrt{1-\sum a_k/n}}=\sqrt{n}\frac{\sum_k a_k}{\sqrt{n-1}}=\sqrt{n} \frac{\sqrt{\sum_k a_k}}{\sqrt{n-1}}.~~~~(1)$$ $\sum_ka_k=1$ is used partially in above. Next using RMS-AM inequality: $\sqrt{\sum_k a_k} \ge \sum_k \sqrt{a_k}/\sqrt{n}$ in (1), we finally get $$S=\sum_{k=1}^n \frac{a_k}{\sqrt{1-a_k}}\ge \frac{\sum_k \sqrt{a_k}}{\sqrt{n-1}}$$
Note: If f''(x)>0 for $x\in D$, then for all $x_k$ in this domain, the Jensen's inequality is $$\frac{1}{n}\sum_{k} f(x_k) \ge f(\sum_k\frac{ x_k}{n})$$