My Math Forum  

Go Back   My Math Forum > College Math Forum > Abstract Algebra

Abstract Algebra Abstract Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
December 6th, 2016, 12:43 PM   #1
Newbie
 
Joined: Dec 2016
From: Vietnam

Posts: 2
Thanks: 0

Condition of inequation

I have two sets of numbers satisfying:
x1+x2+...+xn>=y1+y2+...+yn
Whether it is the case:
x1^2+x2^2+...+xn^2>=y1^2+y2^2+...+yn^2 (xi^2 presents the square of xi).
xi,yi here are not negative numbers.
vdcg15 is offline  
 
December 6th, 2016, 02:40 PM   #2
Global Moderator
 
Joined: May 2007

Posts: 6,276
Thanks: 516

Proof by mathematical induction. Assuming you want to prove it for all n.

x1>y1 implies x1^2>y1^2
x2>y2 implies x2^2>y2^2

Add together and get x1+x2>y1+y2 implies x1^2+x2^2>y1^2+y2^2
mathman is offline  
December 7th, 2016, 01:45 AM   #3
Newbie
 
Joined: Dec 2016
From: Vietnam

Posts: 2
Thanks: 0

What do you think about:
3+1>=2+2
3^2+1^2>=2^2+2^2
vdcg15 is offline  
December 7th, 2016, 01:45 PM   #4
Global Moderator
 
Joined: May 2007

Posts: 6,276
Thanks: 516

2+2>3+0
4+4<9+0
mathman is offline  
Reply

  My Math Forum > College Math Forum > Abstract Algebra

Tags
condition, inequation



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Inequation Dang Elementary Math 5 December 30th, 2013 03:58 PM
inequation metalari Algebra 1 December 2nd, 2012 02:47 AM
Inequation msambinelli Applied Math 5 April 1st, 2012 07:48 AM
inequation kapital Elementary Math 6 December 29th, 2010 10:39 PM
Inequation Alvy Calculus 3 March 27th, 2010 07:32 AM





Copyright © 2017 My Math Forum. All rights reserved.