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 \[a = b\]

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