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

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

Sanjeet Raria - 5 years, 4 months ago

Great @megh choksi Keep it up bro

Deepanshu Gupta - 5 years, 4 months ago

How to mention someone on brilliant?

U Z - 5 years, 4 months ago

Just write @Name which You want to post

Deepanshu Gupta - 5 years, 4 months ago

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

Samuraiwarm Tsunayoshi - 5 years, 4 months ago

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 - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

A wonderful Proof of A.M≥G.M A.M \geq G.MA.M≥G.M inequality

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A wonderful Proof of $A.M \geq G.M$ inequality