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Richard Smith
Poster

Post Number: 24
Posted on Wednesday, 07 November, 2007 - 05:47 pm:   

Hi,

I just wanted to ask about the main methods of proving inequalities. I came across a relatively easy one the other day which was

By considering (x-y)^2, prove that (x^2)+(y^2)>=2xy

I proved this inequality without much trouble assuming the opposite i.e (x^2)+(y^2)<=2xy and rearranging the inequality to form the equation:

(x^2)-2xy+(y^2)<=0

This factorised to become

(x-y)^2<=0

Because there is no real number that can be squared to give a negative number, this was a contradiction and this proves that

(x^2)+(y^2)>=2xy

However, I was considering harder inequalities with x^3, y^3 and higher powers. How are you meant to prove these inequalities without spotting a clever factorisation. Are there are any easy methods to do this?? I have heard of something called Muirhead's inequality but I only partially understood it. Can someone tell me the main methods of proving harder inequalities apart from spotting a clever factorisation.
Ruth Franklin
Prolific poster

Post Number: 380
Posted on Wednesday, 07 November, 2007 - 06:45 pm:   

(For the inequality in your post, it would probably easier to do it just by saying (x-y)2³0, expanding out and adding 2xy to both sides.)
From that inequality, you can divide both sides by 2 and replace x2 and y2 by a=x2 and b=y2 to get
(a+b)/2³Ö(ab)
which is a case of an inequality called the AM-GM inequality (the arithmetic mean of some non-negative numbers is greater than their geometric mean)
In general, the AM-GM inequality says that for n numbers a1,a2,...,an³0
(a1+a2+...+an)/n³(a1a2...an)1/n
This can be useful for proving many inequalities

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