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A wonderful Proof of \[ A.M \geq G.M\] inequality

Lets us consider a semicircle with diameter AB. If C is any point on the semicircle, then ABC forms a right angled triangle with right angle at C .

Let CD be the perpendicular to AB dropped from C . Let AD= a and DB=b

The trianges DAC and DCB are similar , hence

\[\frac{DA}{DC} = \frac{DC}{DB}\]

therefore

\[ DC^{2} = DA . DB = ab\]

and thus \[DC = \sqrt{ab}\]

if r is the radius of the semicircle , then \[DC \leq r\]

we know \[ r = \frac{diameter}{2} = \frac{a + b}{2}\]

Thus

\[\boxed{ \frac{a + b}{2} \geq \sqrt{ab}}\]

If \[DC = r\] that is if the Point D comes on the center of the circle then \[a = b\]

Please reshare if you liked it

Note by U Z
3 years, 1 month ago

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Geometry proves AM-GM. That's the beauty of mathematics. Reshared!

Sanjeet Raria - 3 years, 1 month ago

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Great @megh choksi Keep it up bro

Deepanshu Gupta - 3 years, 1 month ago

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How to mention someone on brilliant?

U Z - 3 years, 1 month ago

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Just write @Name which You want to post

Deepanshu Gupta - 3 years, 1 month ago

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Geometry can also prove RMS, AM, GM, HM too. Look at here.

Samuraiwarm Tsunayoshi - 3 years, 1 month ago

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the only condition when DC = r will be when D is the centre of circle. And taking D as the centre of circle will it self make AD = a = radius = b = DB, why do we even need such a long calculation?

Ahmed Yaseen - 3 years, 1 month ago

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