# $\frac{a_1}{\sqrt{1-a_1}}+\frac{a_2}{\sqrt{1-a_2}}+..+ \frac{a_n}{\sqrt{1-a_n}} \geq \frac{1}{\sqrt{n-1}}(\sqrt{a_1}+\sqrt{a_2}+\dots+ \sqrt{a_n})$

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

• Shouldn't it be $(n-1)S\ge \frac{a_2+a_3+...a_n}{\sqrt{1-a_1}}+\frac{a_1+a_3+...a_n}{\sqrt{1-a_2}}+\dots +\frac{a_2+a_3+...a_{n-1}}{\sqrt{1-a_n}}=\sqrt{1-a_1}+\sqrt{1-a_2}+\cdots +\sqrt{1-a_n}$ Commented Jun 8, 2021 at 16:50
• but if it is $\sqrt{1-a_1}+\sqrt{1-a_2}+\dots+\sqrt{1-a_n}$ i don´t see how get the desired inequality Commented Jun 8, 2021 at 17:04
• HInt: Show that $LHS \geq \sqrt{ n / (n-1 ) } \geq RHS$. Jensens works directly. CS works too, with a bit more creativity. Commented Jun 8, 2021 at 17:39
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 $$\frac{1}{n}\sum_{k} f(x_k) \ge f(\sum_k\frac{ x_k}{n})$$