If $a_1,a_2,\dots ,a_n\ge 0$ such that $a_1+a_2+\dots a_n=1$ show that $\frac{a_1}{\sqrt{1-a_1}}+\frac{a_2}{\sqrt{1-a_2}}+\dots +\frac{a_n}{\sqrt{1-a_n}} \geq \frac{1}{\sqrt{n-1}}(\sqrt{a_1}+\sqrt{a_2}+\dots+ \sqrt{a_n})$.


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

$$S\geq \frac{a_2}{\sqrt{1-a_1}}+\frac{a_3}{\sqrt{1-a_2}}+\dots +\frac{a_1}{\sqrt{1-a_n}}$$

$$S \geq \frac{a_3}{\sqrt{1-a_1}}+\frac{a_4}{\sqrt{1-a_2}}+\dots +\frac{a_2}{\sqrt{1-a_n}}$$


$$S \geq \frac{a_n}{\sqrt{1-a_1}}+\frac{a_{n-1}}{\sqrt{1-a_2}}+\dots +\frac{a_1}{\sqrt{1-a_1}}$$

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.

  • $\begingroup$ Shouldn't it be $(n-1)S\ge \frac{a_2+a_3+...a_n}{\sqrt{1-a_1}}+\frac{a_1+a_3+...a_n}{\sqrt{1-a_2}}+\dots +\frac{a_2+a_3+...a_{n-1}}{\sqrt{1-a_n}}=\sqrt{1-a_1}+\sqrt{1-a_2}+\cdots +\sqrt{1-a_n}$ $\endgroup$
    – Asher2211
    Commented Jun 8, 2021 at 16:50
  • $\begingroup$ Ups, i get a mistake $\endgroup$
    – Juan T
    Commented Jun 8, 2021 at 17:03
  • $\begingroup$ 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 $\endgroup$
    – Juan T
    Commented Jun 8, 2021 at 17:04
  • 1
    $\begingroup$ HInt: Show that $ LHS \geq \sqrt{ n / (n-1 ) } \geq RHS$. Jensens works directly. CS works too, with a bit more creativity. $\endgroup$
    – Calvin Lin
    Commented Jun 8, 2021 at 17:39
  • $\begingroup$ @CalvinLin please illustrate how Cauchy Schwarz will work $\endgroup$ Commented Jun 8, 2021 at 18:13

1 Answer 1


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 $$\frac{1}{n}\sum_{k} f(x_k) \ge f(\sum_k\frac{ x_k}{n})$$


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