For positive distinct (x,y): [ (\sqrt{x}-\sqrt{y})^2>0\implies x+y>2\sqrt{xy}\implies\dfrac{x+y}{2}>\sqrt{xy} ]
i.e. A.M. > G.M.
Proved.
11TH•Mathematics•Sequences And Series•Question 3
Overview
If both $x$ and $y$ are positive distinct real numbers, show that the geometric mean between $x$ and $y$ is less than their arithmetic mean.
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