Monday, December 29, 2014

Arithmetic Geometric Series

This is a famous inequality that one should know. The case for two numbers is really easy $(\sqrt{a}-\sqrt{b})^2 \ge 0 \Rightarrow a+b \ge 2\sqrt{a}\sqrt{b} \Rightarrow \frac{a+b}{2} \ge \sqrt{ab}$
The case for 4 variables is also pretty straight forward $\dfrac{a+b+c+d}{4} =\dfrac{ \dfrac{a+b}{2} +\dfrac{c+d}{2}}{2} \ge \dfrac{\sqrt{ab}+\sqrt{cd}}{2} \ge \sqrt{ \sqrt{ab} \sqrt{cd}} \ge \sqrt[4]{abcd}$
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 $d = \sqrt[3]{abc}$. Then $\dfrac{a+b+c+d}{4} \ge \sqrt[4]{abcd} =\sqrt[4]{abc(abc)^{\frac{1}{3}}} = \sqrt[4]{(abc)^{\frac{4}{3}}} =(abc)^{\frac{1}{3}} =d$ and so we have $\dfrac{a+b+c+d}{4} \ge d \Rightarrow \dfrac{a+b+c}{4}+\dfrac{d}{4} =d \Rightarrow \dfrac{a+b+c}{4} = \dfrac{3d}{4} \Rightarrow \dfrac{a+b+c}{3} \ge d \Rightarrow \dfrac{a+b+c}{3} \ge \sqrt[3]{abc}$



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