Arithmetic Geometric Series
This is a famous inequality that one should know. The case for two numbers is really easy (√a−√b)2≥0⇒a+b≥2√a√b⇒a+b2≥√ab
The case for 4 variables is also pretty straight forward a+b+c+d4=a+b2+c+d22≥√ab+√cd2≥√√ab√cd≥4√abcd
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=3√abc. Then a+b+c+d4≥4√abcd=4√abc(abc)13=4√(abc)43=(abc)13=d and so we have a+b+c+d4≥d⇒a+b+c4+d4=d⇒a+b+c4=3d4⇒a+b+c3≥d⇒a+b+c3≥3√abc
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