Let $x_1,x_2,\cdots,x_n \in(0,1)$ , $n\geq2$. Prove that$$\frac{\sqrt{1-x_1}}{x_1}+\frac{\sqrt{1-x_2}}{x_2}+\cdots+\frac{\sqrt{1-x_n}}{x_n}<\frac{\sqrt{n-1}}{x_1 x_2 \cdots x_n}.$$
Problem
Source: China Shenzhen ,13 Aug 2015
Tags: Inequality, algebra, inequalities, China
YlOwOlY
13.08.2015 11:29
I would like to use mathematical induction for proving this inequality.
At first, we need to confirm that the statement is true for $ n=2 $
\[ \frac{\sqrt{1-x}}{x}+\frac{\sqrt{1-y}}{y} < \frac{1}{xy}, \forall x,y \in \left(0,1\right) \]
\[ \Leftrightarrow y\sqrt{1-x}+x\sqrt{1-y} < 1 \]
\[ \Leftrightarrow 2xy\sqrt{1-x-y+xy} < 1-x-y+2xy \]
However, by using AM-GM inequality, we have
\[ RHS=\left(1-x-y+xy\right)+xy > \left(1-x-y+xy\right)+x^2 y^2 \ge LHS \]
Suppose it’s true for $ n=m $ and we should prove that it’s still right for $ n=m+1 $
WLOG $ k=a_{1}=\max\{a_{1}, a_{2}, … , a_{m+1}\} $, then by the hypothesis,
\[ LHS < \frac{\sqrt{1-k}}{k}+\frac{\sqrt{m-1}}{\prod\limits_{i \neq 1}{a_{i}}} < \frac{\sqrt{1-k}}{k}+\frac{\sqrt{m-1}}{\prod{a_{i}}} \]
And thus, it suffices to show the following inequality
\[ \frac{\sqrt{1-k}}{k} \le \frac{\sqrt{m}-\sqrt{m-1}}{\prod{a_{i}}} \]
Here, I’ll prove a stronger one. Precisely, that is
\[ \frac{\sqrt{1-k}}{k} \le \frac{\sqrt{m}-\sqrt{m-1}}{k^{m+1}} \]
\[ \Leftrightarrow \left(\sqrt{m}+\sqrt{m-1}\right)\sqrt{1-k} \le \frac{1}{k^m} \]
Note that we could find a constant $ \alpha $ so that $ 1+\alpha=\frac{1}{k} $
Then it’s equivalent to $ \left(\sqrt{m}+\sqrt{m-1}\right)\sqrt{\alpha} \le \left(1+\alpha\right)^{m-\frac{1}{2}} $
After using Bernoulli and AM-GM inequality, we have
\[ \left(1+\alpha\right)^{m-\frac{1}{2}} > 1+\left(m-\frac{1}{2}\right)\sqrt{\alpha} \ge 2\sqrt{\left(m-\frac{1}{2}\right)\alpha}>\left(\sqrt{m}+\sqrt{m-1}\right)\sqrt{\alpha} \]
Notice that $ f\left(x\right)=\sqrt{x} $ is a concave function.
sqing
14.08.2015 05:37
sqing wrote: Let $x_1,x_2,\cdots,x_n \in(0,1)$ , $n\geq2$. Prove that$$\frac{\sqrt{1-x_1}}{x_1}+\frac{\sqrt{1-x_2}}{x_2}+\cdots+\frac{\sqrt{1-x_n}}{x_n}<\frac{\sqrt{n-1}}{x_1 x_2 \cdots x_n}.$$ $n=2:$ $\frac{\sqrt{1-x_1}}{x_1}+\frac{\sqrt{1-x_2}}{x_2}<\frac{1}{x_1x_2}\left(x_2\sqrt{1-x^2_1}+x_2\sqrt{1-x^2_1}\right)\leq\frac{1}{x_1x_2}\left(\frac{x^2_2+1-x^2_1}{2}+ \frac{x^2_1+1-x^2_2}{2}\right)=\frac{1}{x_1x_2}.$
sqing
14.08.2015 15:31
sqing wrote: Let $x_1,x_2,\cdots,x_n \in(0,1)$ , $n\geq2$. Prove that$$\frac{\sqrt{1-x_1}}{x_1}+\frac{\sqrt{1-x_2}}{x_2}+\cdots+\frac{\sqrt{1-x_n}}{x_n}<\frac{\sqrt{n-1}}{x_1 x_2 \cdots x_n}.$$ The problem is equivalent to Let $x_1,x_2,\cdots,x_n >0$ , $n\geq2$. Prove that$$\sqrt{x_1(x_1+1)}+\sqrt{x_2(x_2+1)}+\cdots+ \sqrt{x_n(x_n+1)}<\sqrt{n-1}\cdot(x_1+1)( x_2+1) \cdots (x_n+1).$$
lemon94
25.07.2019 21:48
YlOwOlY wrote:
Here, I’ll prove a stronger one. Precisely, that is
\[ \frac{\sqrt{1-k}}{k} \le \frac{\sqrt{m}-\sqrt{m-1}}{k^{m+1}} \]
Where did you get $k^{m+1}$ in the denominator?
VicKmath7
05.09.2021 16:26
Any other solution? I think in #2 Bernoulli's ineq was applied in the reverse direction.
HamstPan38825
07.01.2025 01:47
This is fairly straightforward with induction. For the $n=2$ case, \[\left(a\sqrt{1-b} + b\sqrt{1-a}\right)^2 < \left(a^2+1-a\right)\left(b^2+1-b\right) < 1\]by Cauchy-Schwarz. It turns out the inductive step works the same way: \begin{align*} \sum_{i=1}^n \frac{\sqrt{1-x_i}}{x_i} &< \frac{\sqrt{n-2}}{x_1x_2 \cdots x_{n-1}} + \frac{\sqrt{1-x_n}}{x_n} \\ &= \frac{x_n\sqrt{n-2}+x_1x_2\cdots x_{n-1}\sqrt{1-x_n}}{x_1x_2 \cdots x_n} \\ &\leq \frac{\sqrt{\left(x_n^2+1-x_n\right)\left(n-2+(x_1x_2 \cdots x_{n-1})^2\right)}}{x_1x_2 \cdots x_n} \\ &< \frac{\sqrt{n-1}}{x_1x_2 \cdots x_n} \end{align*}as $x_1 x_2 \cdots x_{n-1} < 1$ by assumption.