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December 6th, 2016, 01:43 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. 
December 6th, 2016, 03:40 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,214 Thanks: 492 
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 
December 7th, 2016, 02:45 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 
December 7th, 2016, 02:45 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,214 Thanks: 492 
2+2>3+0 4+4<9+0 

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