Suppose $a$ and $b$ are real numbers such that $0 < a < b < 1$. Let $$x= \frac{1}{\sqrt{b}} - \frac{1}{\sqrt{b+a}},\hspace{1cm} y= \frac{1}{b-a} - \frac{1}{b}\hspace{0.5cm}\textrm{and}\hspace{0.5cm} z= \frac{1}{\sqrt{b-a}} - \frac{1}{\sqrt{b}}.$$ Show that $x$, $y$, $z$ are always ordered from smallest to largest in the same way, regardless of the choice of $a$ and $b$. Find this order among $x$, $y$, $z$.
Problem
Source: 2nd National Women's Contest of Mexican Mathematics Olympiad 2023 , level 1+2 p7
Tags: Mexico, algebra, inequalities
AlanLG
24.07.2023 00:01
We claim that $y>z>x$ Lemma 1: $y>z$ Proof: Let $f(x)=\frac{1}{x}-\frac{1}{\sqrt{x}}$, note that $f^\prime (x)=\frac{\sqrt{x}-2}{2x^2}<0, \forall x\in(0, 4)$ therefore $f(x)$ is decreasing in that interval, as $1>b>b-a>0$ then $f(b-a)=\frac{1}{b-a}-\frac{1}{\sqrt{b-a}}>\frac{1}{b}-\frac{1}{\sqrt{b}}=f(b)$ $\hspace{20cm}\blacksquare$ Lemma 1: $z>x$ Proof: As $g(x)=\frac{1}{\sqrt{x}}$ is convex for $x>0$, by Jensen $$\frac{g(b-a)+g(b+a)}{2}>g\left(\frac{(b-a)+(b+a)}{2}\right) \iff$$$$\frac{\frac{1}{\sqrt{b-a}}+\frac{1}{\sqrt{b+a}}}{2}>\frac{1}{\sqrt{\frac{(b-a)+(b+a)}{2}}}=\frac{1}{\sqrt{b}}$$$\hspace{20cm}\blacksquare$
RagvaloD
24.07.2023 00:12
$y>z$ because $y=z(\frac{1}{\sqrt{b-a}}+\frac{1}{\sqrt{b}})>z$ $z>x$ because $z^2 =\frac{1}{b-a}+\frac{1}{b+a}+\frac{2}{\sqrt{b^2-a^2}}=\frac{2b}{b^2-a^2}+\frac{2}{\sqrt{b^2-a^2}}>\frac{4}{b}=x^2$
sqing
24.07.2023 03:51
RagvaloD wrote:
$y>z$ because $y=z(\frac{1}{\sqrt{b-a}}+\frac{1}{\sqrt{b}})>z$
$z>x$ because $z^2 =\frac{1}{b-a}+\frac{1}{b+a}+\frac{2}{\sqrt{b^2-a^2}}=\frac{2b}{b^2-a^2}+\frac{2}{\sqrt{b^2-a^2}}>\frac{4}{b}=x^2$
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