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This question already has an answer here:

If $A.M$ and $G.M$ are Arithmetic Mean And Geometric Mean respectively then prove that $A.M \ge G.M$.

My Attempt :

Let $a$ and $b$ are any two real positive numbers. Then:
$$A.M=\frac{a+b}{2}$$

$$G.M =\sqrt{ab} $$

Now how can we show that $A.M.\ge G.M.$ ? Please help me..

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marked as duplicate by Yuriy S, colormegone, Ramiro, Daniel W. Farlow, MCT Jun 3 '16 at 23:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$$0\leq\left(\sqrt{a}-\sqrt{b}\right)^{2}=a+b-2\sqrt{ab}.$$

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Let $AF=a, FB=b$. We construct the semicircle with diameter $AB$. Then $$DE=\frac{a+b}2; CF=\sqrt{ab}$$ $$DE\ge CF \Rightarrow \frac{a+b}2\ge\sqrt{ab}$$

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Extending Marco Cantarini's answer $$0\leq\left(\sqrt{a}-\sqrt{b}\right)^{2} =a+b-2\sqrt{ab} => 0\leq\a+b-2\sqrt{ab} => 2\sqrt{ab}\leq\a+b => \sqrt{ab}\leq\frac{a+b}{2} => G.M.\leq\A.M.$$

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