It is currently Sun Jul 22, 2018 7:01 pm

 All times are UTC [ DST ]

 Page 1 of 1 [ 3 posts ]
 Print view Previous topic | Next topic
Author Message
 Post subject: InequalityPosted: Sun Aug 13, 2017 4:55 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 146
Location: Melbourne, Australia
Let $x, y,z >0$ satisfying $x+y+z=1$. Prove that

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}$

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

Top

 Post subject: Re: InequalityPosted: Mon Aug 14, 2017 9:47 am
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Hi Riemann.

It is sufficient to prove that

$\displaystyle{a+b+c\geq \sqrt{3\,a\,b\,c}}$, where $\displaystyle{a\,,b\,,c>1}$ and $\displaystyle{a\,b+b\,c+c\,a=a\,b\,c}$.

So,

\displaystyle{\begin{aligned}a+b+c\geq \sqrt{3\,a\,b\,c}&\iff (a+b+c)^2\geq 3\,a\,b\,c\\&\iff (a^2+b^2+c^2)+2\,(a\,b+b\,c+c\,a)-3\,(a\,b+b\,c+c\,a)\geq 0\\&\iff (a^2+b^2+c^2)-(a\,b+b\,c+c\,a)\geq 0\\&\iff 2\,a^2+2\,b^2+2\,c^2-2\,a\,b-2\,b\,c-2\,c\,a\geq 0\\&\iff (a-b)^2+(b-c)^2+(c-a)^2\geq 0 \end{aligned}}

and the last one is true. The equality holds, if, and only, if, $\displaystyle{a=b=c}$ and then

$\displaystyle{a^2+a^2+a^2=a^3\iff a^3=3\,a^2\iff a=3=b=c}$.

We conclude that, if $\displaystyle{x\,,y\,,z>0}$ such that $\displaystyle{x+y+z=1}$, then

$\displaystyle{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\geq \sqrt{\dfrac{3}{x\,y\,z}}}$

and the equality holds if, and only if, $\displaystyle{\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}=3}$

or equivalently, if, and only if, $\displaystyle{x=y=z=\dfrac{1}{3}}$.

Top

 Post subject: Re: InequalityPosted: Sun Aug 20, 2017 9:25 am

Joined: Sat Nov 14, 2015 6:32 am
Posts: 146
Location: Melbourne, Australia
Thank you Papapetros Vaggelis. My solution is as follows.

Since $\frac{1}{x} \; , \; \frac{1}{y} \; , \; \frac{1}{z} >0$ then the numbers

$\sqrt{\frac{1}{x} + \frac{1}{y}} \; , \; \sqrt{\frac{1}{x} +\frac{1}{z}} \; , \; \sqrt{\frac{1}{y} + \frac{1}{z}}$

could be sides of a triangle. The area of this triangle is

$\mathcal{A} = \frac{1}{2} \sqrt{\frac{1}{xy} + \frac{1}{xz} + \frac{1}{yz}} = \frac{1}{2} \sqrt{\frac{x+z+y}{xyz}} = \frac{1}{2\sqrt{xyz}}$

However , in any triangle is holds that [WeitzenbĂ¶ck]

\begin{equation*} a^2+b^2+c^2 \geq 4 \mathcal{A} \sqrt{3} \end{equation*}

where $\mathcal{A}$ is the area of the triangle. Thus

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}$

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 3 posts ]

 All times are UTC [ DST ]

#### Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

Search for:
 Jump to:  Select a forum ------------------ Algebra    Linear Algebra    Algebraic Structures    Homological Algebra Analysis    Real Analysis    Complex Analysis    Calculus    Multivariate Calculus    Functional Analysis    Measure and Integration Theory Geometry    Euclidean Geometry    Analytic Geometry    Projective Geometry, Solid Geometry    Differential Geometry Topology    General Topology    Algebraic Topology Category theory Algebraic Geometry Number theory Differential Equations    ODE    PDE Probability & Statistics Combinatorics General Mathematics Foundation Competitions Archives LaTeX    LaTeX & Mathjax    LaTeX code testings Meta