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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 (xy)^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 (xy)^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 (xy)^{2}³0, expanding out and adding 2xy to both sides.) From that inequality, you can divide both sides by 2 and replace x^{2} and y^{2} by a=x^{2} and b=y^{2} to get (a+b)/2³Ö(ab) which is a case of an inequality called the AMGM inequality (the arithmetic mean of some nonnegative numbers is greater than their geometric mean) In general, the AMGM inequality says that for n numbers a_{1},a_{2},...,a_{n}³0 (a_{1}+a_{2}+...+a_{n})/n³(a_{1}a_{2}...a_{n})^{1/n} This can be useful for proving many inequalities 
