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Monday, December 29, 2014

Arithmetic Geometric Series

This is a famous inequality that one should know. The case for two numbers is really easy (ab)20a+b2aba+b2ab
The case for 4 variables is also pretty straight forward a+b+c+d4=a+b2+c+d22ab+cd2abcd4abcd
The case for 3 variables is a little bit tricky but can be proved quite easily if we use use the 4 variable proof. Let d=3abc. Then a+b+c+d44abcd=4abc(abc)13=4(abc)43=(abc)13=d and so we have a+b+c+d4da+b+c4+d4=da+b+c4=3d4a+b+c3da+b+c33abc

 

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