My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Thanks Tree3Thanks
  • 1 Post By CRGreathouse
  • 1 Post By mathman
  • 1 Post By Olinguito
Reply
 
LinkBack Thread Tools Display Modes
April 11th, 2014, 09:05 AM   #1
Senior Member
 
Joined: Nov 2013

Posts: 434
Thanks: 8

show

Show that if real numbers x,y,z>0 satisfy the relation

xyz+xy+yz+zx=4,
then x+y+z>= 3.
mared is offline  
 
April 11th, 2014, 09:20 AM   #2
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,681
Thanks: 2659

Math Focus: Mainly analysis and algebra
The first equation gives us $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3$$
Which I think turns out to be useful in this one.
v8archie is offline  
April 11th, 2014, 09:34 AM   #3
Global Moderator
 
CRGreathouse's Avatar
 
Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
That doesn't look quite right...
Thanks from v8archie
CRGreathouse is offline  
April 11th, 2014, 12:22 PM   #4
Global Moderator
 
Joined: May 2007

Posts: 6,821
Thanks: 723

Quote:
Originally Posted by v8archie View Post
The first equation gives us $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3$$
Which I think turns out to be useful in this one.
Right side is wrong. Should be 4/(xyz) - 1
Thanks from v8archie
mathman is offline  
April 11th, 2014, 01:27 PM   #5
Senior Member
 
Olinguito's Avatar
 
Joined: Apr 2014
From: Greater London, England, UK

Posts: 320
Thanks: 156

Math Focus: Abstract algebra
Let $t=x+y+z$.

By AM–GM,
$$t^3\ =\ (x+y+z)^3\ \geqslant\ 27xyz\ \ldots\fbox1$$
By Cauchy–Schwarz,
$$\begin{align*}
t^2\ =\ (x+y+z)^2 &\leqslant\ (x^2+y^2+z^2)(1^2+1^2+1^2)
\\\\ &=\ 3(x^2+y^2+z^2)
\\\\ &=\ 3t^2-6(xy+yz+zx)\end{align*}$$
$\displaystyle \therefore\ t^2\ \geqslant\ 3(xy+yz+zx)\ =\ 12-3xyz\ \ldots\fbox2$

$\displaystyle \fbox1+9\times\fbox2$ gives
$$\begin{align*}
t^3+3t^2-108 &\geqslant\ 0
\\\\ (t-3)(t+6)^2 &\geqslant\ 0
\end{align*}$$
Since $(t+6)^2>0$, it follows that $t-3\geqslant0$. QED
Thanks from v8archie
Olinguito is offline  
April 11th, 2014, 05:32 PM   #6
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,681
Thanks: 2659

Math Focus: Mainly analysis and algebra
Oops. I should answer when I have a bit of time to think more clearly!
v8archie is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
show



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Show that 5^x. ? Shamieh Algebra 5 July 3rd, 2013 06:35 PM
show that [1/((r + 1)!)] = [1/(r!)] - [1/((r+1)!)] the_liong Algebra 1 September 12th, 2010 12:46 AM
want to show that show that two infinite summations R equal notnaeem Real Analysis 4 August 16th, 2010 12:32 PM
show that f is zero john235 Real Analysis 2 March 2nd, 2009 07:48 AM
How do I show that f(c)=0 for some c in (0,1) peka0027 Real Analysis 2 February 26th, 2009 06:01 PM





Copyright © 2019 My Math Forum. All rights reserved.