Arithmetic Mean And Geometric Mean

Question

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..

See the answer to this question http://math.stackexchange.com/questions/27328/geometric-mean-never-exceeds-arithmetic-mean?rq=1 — user_of_math, Jun 03 '16 at 17:25

score 2 · Answer 1 · answered Jun 03 '16 at 17:22

2

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

answered Jun 03 '16 at 17:22

Marco Cantarini

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could you please elaborate a bit more? I can't understand – user335710 Jun 03 '16 at 17:24
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@user335710 A squared number is always greater or equal than $0$. About the equality see https://en.wikipedia.org/wiki/Binomial_(polynomial) – Marco Cantarini Jun 03 '16 at 17:28

score 2 · Answer 2 · answered Jun 03 '16 at 18:02

2

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

answered Jun 03 '16 at 18:02

Roman83

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score 0 · Answer 3 · edited Jan 06 '20 at 05:13

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

edited Jan 06 '20 at 05:13

Martin Sleziak

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answered Jun 03 '16 at 17:49

Shona

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Arithmetic Mean And Geometric Mean

3 Answers3