5
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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 Jun 8 at 16:50
  • $\begingroup$ Ups, i get a mistake $\endgroup$ – Juan T Jun 8 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 Jun 8 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 Jun 8 at 17:39
  • $\begingroup$ @CalvinLin please illustrate how Cauchy Schwarz will work $\endgroup$ – Lalit Tolani Jun 8 at 18:13
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.