Visual proof every student should see

This is one of my favorite visual proofs that uses geometry to demonstrate an algebraic inequality. The arithmetic mean is always greater than or equal to the geometric mean, with equality if and only if the numbers are equal. (Requires the numbers to be non-negative). #shorts

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[bhavyapal]

This visual proof will make me remember it for the rest of my life. Thanks

[jackhandma]

I love the part where Euclid said "it's chordin' time" and drew a lot of chords.

[pramodpoddar]

Another method to prove AM>=GM is as follows
a and b are two positive numbers
(a-b)²>=0
a²+b²-2ab>=0
Add 4ab to both sides, since a and b are positive, inequality doesn't change
a²+b²+2ab>=4ab
Both sides are positive so we can take square root without changing inequality
(a+b)>=2√ab
(a+b)/2>=√ab

[jacobsuszczynski]

Never seen this before, great visualization

[ppg]

Coming from studying in a South American country, my german teacher showed us this in one of the first classes, as common knowledge, I was blowned away lmao

[barbossablink]

Woahh that's amazing never really thought in this way !

[IamAbhinav]

Your videos are very informative and interesting

[drainedzombie]

I miss my 15 &16s where I had bunch of such visualisations regarding proofs in co-ordinate geometry while preparing for JEE which I couldn't document. I'm 22 now

[bioengboi]

the AM is the first root of x^1 + y^1 with a weight of 0.5, the weight shifts to 0.333... with say z^1. While the GM is the zeroth root of x^0 + y^0 with each weighing 0.5 since 0 < 1 the GM < AM. take this further with the harmonic mean: the negative first root of x^-1 + y^-1 with each weighing 0.5 since -1 < 0, the HM < GM < AM. :)

[adamrezabek]

When we were doing inequalities in the 9th grade, I remember seeing tons of proofs for general inequalities witch included x+y>2√xy. Like AM-GM, Minkowski or Cauchy-Schwarz. It kinda seams like every inequality is just more general case of this

[uchirrod]

I remember my teacher used this same visual proof to prove the quadratic and harmonic means inequalities as well, but i can't remember how. If you happen to find out anything about it, would be cool if you made a video on that as well!

[onionbroisbestwaifu]

You can also visualize this with the Cauchy-Schwartz

[UltraMagnus]

Thanks for this Sir please make more such type of videos

[alexismiller]

I find it crazy how this is not thought along side the AM-GM mean usually.
I got lucky and saw this first and have never forgot it since

[erikev]

The visual proof is fine, but this fact is really a trivial matter: A) it is intuitive and obvious since the derivative of the exponential is >0 for all x. and B) just perfecting the square trivially reduces this to (x-y)²≥0 (all square numbers are positive)

[veenadayma]

Yep, this was the exact geometric visualisation that popped in my mind a few weeks after I saw this theorem.

A VERY SIMILAR visualisation consists of imagining two circles with centres C1 and C2 that touch each other externally. Suppose their radii are x and y respectively. Draw a common tangent to both of them. It will meet the circles at a point each. Join each of the points with their respective centres and also join the two centres C1 and C2. You get a right trapezium.

To find the length of this common tangent, say l, draw a line parallel to the tangent from the centre of the smaller circle, meeting the radius of the larger circle. The new four-sided figure so-formed is a parallelogram and hence the length of this parallel line segment is l.


Now, in the remaining right angled triangle, we know that the length of the hypotenuse, i.e., x+y, is always greater than its individual side.

So, l ≤ x+y.

The third side of this triangle is of length x-y, so from Pythagoras Theorem,

(x-y)² + l² = (x+y)²

l² = 4xy or l = 2√xy.
But, l ≤ x+y. Therefore,

2√xy ≤ x+y or
(x+y)/2 ≥ √xy

So, A.M. of two quantities is always greater than their G.M. The equality holds when x=y, in which case the right trapezium reduces to a rectangle and hence l=2x=2y.

This completes another proof. Although the one you showed is generally much more widely preferred, this one also helps you prove that the length of the common tangent of two circles touching each other externally is 2√xy.

[jayeshkumar]

This is cool. But, algebraic path is also very easy and clear.
(x-y)^2 > 0 always
So, x^2-2xy+y^2 > 0
So x^2+y^2 > 2xy
So (x^2+y^2)/2 > xy ie Sq Rt of x^2•y^2
[for non negative distinct numbers]
Thus, (A+B)/2 > Sq Root of A•B
THAT'S
AM > GM For Distinct non negative numbers


Note - U can't take numbers set like 4, 4, -4, -4.

Here, AM is 0. GM is 4.


Rule applies to non negative distinct numbers (any number of numbers).

[last_fantasy]

The thing that was actually needed the visualization is g.g=x.y

[reimannintegrability]

When Pythagoras said " It's Circalian time", Euclid drew circles to prove AM and GM

[MrRenanwill]

(x-y)^2>=0
Hence
x^2-2xy+y^2 >= 0
Thus,
x^2+2xy+y^2>=4xy
Finally,
(x+y)^2>=(2xy)^2, which derives the result.