Problem-Solving Trick No One Taught You: RMS-AM-GM-HM Inequality

Рет қаралды 224,843

Күн бұрын

This inequality is famous in math competitions and in theoretical proofs. But why is it true? The video presents a great geometric visualization and proof for two variables. Pay attention--I'll use this inequality in an upcoming video!
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Пікірлер: 432

@morethejamesx39 6 жыл бұрын

Hey this is Presh Talkwalker

@JonSebastianF 6 жыл бұрын

Identity crises can be brief but hard-hitting...

@Tehom1 6 жыл бұрын

Because somebody in the comments misheard his name last video, I think.

@ffggddss 6 жыл бұрын

Nah, people have been mis-hearing his name for ages & ages. He's just getting slower. Or something. Incidentally, you misspelled his name. There's no "e" in it. It ends with "ar." Fred

@WardenclyffeResearch 6 жыл бұрын

Did you figure this out?

@darreljones8645 6 жыл бұрын

Since "Walker" is a common English-language last name, I'm sure many people thought his name was "Preshtal Walker".

@Jack_Callcott_AU 5 жыл бұрын

I learned this for the first time when I was about 40 yrs old. This geometric proof is just so elegant. Such a shame I never encountered it at school or university.

@billy.7113 6 жыл бұрын

*Thank you for the math lesson.* I've gained much more knowledge by watching this than those debatable puzzles.

@pe3akpe3et99 4 жыл бұрын

first answer

@u5s9e2hb4ijk7bv 4 жыл бұрын

How to get root(ab): Use proportions. Let c be the length of the red segmant. Then a/c = c/b, which implies c^2 = ab, since the triangles are similar.

@raihanmaulana3744 8 ай бұрын

how do you know both triangles are congruent?

@GermansEagle 6 жыл бұрын

Seriously, how have I never heard of this proof.

@fernandowong5799 5 жыл бұрын

because this proof only works for two terms, which isn't the most useful form

@sunilrampuria9339 5 жыл бұрын

@@fernandowong5799 we can then apply induction to prove it for n number of terms.

@brodieenrique1003 3 жыл бұрын

i know it is pretty off topic but does anyone know of a good site to stream newly released movies online?

@kenzorowen2048 3 жыл бұрын

@Brodie Enrique Lately I have been using FlixZone. You can find it by googling :)

@abramdrake4510 3 жыл бұрын

@Kenzo Rowen definitely, I have been using FlixZone for months myself :)

@looney1023 5 жыл бұрын

This is the best video you've made thus far. A cool visualization / geometric proof of a useful theorem. Nice job

@TheOfficialCzex 6 жыл бұрын

Fresh Tall Water. Got it.

@greg939 4 жыл бұрын

You mean Fresh Saltwater

@sahilsagwekar 3 жыл бұрын

Presh tall walker

@sahilsagwekar 3 жыл бұрын

@Sai Sasank presh talwalkar l, his sirname is indian

@aashsyed1277 3 жыл бұрын

@@sahilsagwekar no he lives in USA

@mathlegendno12 3 жыл бұрын

@@sahilsagwekar r/whoosh

@Trinexx42 6 жыл бұрын

I have algebraic proofs of the inequalities using proof by contradiction: First, suppose that RMS

@bhardwajr01 6 жыл бұрын

Nevan Lowe u may just suggest that it will be proven by contradiction and leave it to the readers

@jaroslavsevcik3421 6 жыл бұрын

But he wanted to provide the solution too. It is his right. So next time let your suggestions at home please.

@dorijancirkveni 6 жыл бұрын

Jaroslav Ševčík The joke: y=1-x^2 You: (0,0)

@marcusyang7686 6 жыл бұрын

Jaroslav Ševčík he is obviously just joking. In most math Olympiad books there's always statements like this.

@facitenonvictimarum174 6 жыл бұрын

Nevan Lowe ...Thanks for sharing that with us, a math lesson in itself, and for taking all the time that must have been necessary to present it so well with the math symbol limitations of a computer keyboard. Good job!!

@deadfish3789 6 жыл бұрын

It took me quite a while to work out where you got sqrt(ab) and xAM=GM^2. So you could go into those more explicitly

@adamwho9801 5 жыл бұрын

Similar triangles have sides of equal ratios GM/x = AM/GM

@anishkrishnan9698 3 жыл бұрын

Yes, so from similar triangles from his diagram: h/a = b/h ==> h^2 = ab ==> h = sqrt(ab) = GM

@zeynarz7614 2 жыл бұрын

@@anishkrishnan9698 Thanks a lot!

@parahumour4619 2 жыл бұрын

@@adamwho9801 Aaah that seems easier I took wrote pythogoras equations for three triangles and equated them, 4 steps but yeah works

@GermansEagle 6 жыл бұрын

Thats awesome man! Nice video !!!!

@AnshuKumar-oj8ww 6 жыл бұрын

You have done everything elegantly. Nice immaculate work ! 👍

@koenth2359 6 жыл бұрын

Wow, very amazing and elegant! We can even see other things from the graph, for example that AM/RMS >= HM/GM. Explanation: These ratios are the cosines of the top angles. And the top angles are arctan((b-a)/2GM) and arctan((b-a)/2AM) respectively. Since arctan is ascending and AM and GM are in the denominator, the right top angle is smaller than the left top angle. And since cosine is a descending function on the interval [0, pi/2], the ratio AM/RMS is larger (or equal) than RM/GM.

@jackthatmonkey8994 5 жыл бұрын

My mind gets blown everytime when I watch your stuff. I can barely keep up.

@bernhard5295 6 жыл бұрын

Really nice prove! I would like to see more of this format. Thumps up👍

@popogast 6 жыл бұрын

Most useful contribution of the last weeks. Thank You.

@shanmugasundaram9688 6 жыл бұрын

The video description of all the means merging together when a=b is wonderful.

@Etothe2iPi 6 жыл бұрын

Great idea to pepper your videos from time to time with this kind of educational content!

@notspaso6644 6 жыл бұрын

Great one! Hope to see more videos like these in the future ^_^

@raghavagarwal5435 6 жыл бұрын

Really helpful. Thank you very much. I am a twelth grader and have never seen such an interesting proof of this inequality.

@jokarmaths7771 2 жыл бұрын

amazing ...kzfaq.info/get/bejne/nsqXqdiAt9SpmXU.html

@Epoch11 6 жыл бұрын

A video on why each of these means is useful would be nice. I'm not a mathematician and sure I can go look it up myself, but it would be much easier for me if you did it. Jokes aside, an in depth explanation of these various means would make a video I would definitely watch.

@davebacknolaliki1452 Жыл бұрын

en.m.wikipedia.org/wiki/Mean

@titan1235813 6 жыл бұрын

IMO, this is seriously one of your best videos ever. This one gets to show us that Geometry, I believe, is intrinsically linked to all of Mathematics, even with the most seemingly unrelated topic. What a beautiful proof, Presh. Thank you!

@paridhaxholli 16 күн бұрын

get it, useful in IMO 😂

@isaacpark1016 6 жыл бұрын

Beautifully demonstrated. Love it!

@jokarmaths7771 2 жыл бұрын

amazing ...kzfaq.info/get/bejne/nsqXqdiAt9SpmXU.html

@PhilipBlignaut 6 жыл бұрын

The best description regarding means ever!

@sanseng000 6 жыл бұрын

Simply superb! Super excellent! Awesomely simple.

@ahmedbaig7279 5 жыл бұрын

I should be similar with all these series. Arithmetic Means is used in Statistics. Geometric means is used in calculation of population and compound interest. Wonderfully you have proveded some associations with other two.

@alvarezjulio3800 4 жыл бұрын

What a beauty! That was awesome! Thank Sir!

@iamyoda7917 6 жыл бұрын

Real math! Hooray!

@ankitjain3760 2 жыл бұрын

Me a 36 years old failure in both professional and personal life loves your video try to solve questions, watch them many times. They are lifeline for me.

@user-uo8hc1ju4l 2 жыл бұрын

helpful indeed, a lot better than complicated ways, my goodness, thank you for this super cool way. loved it

@babitamishra524 3 жыл бұрын

I was searching for such geometric approach for proving it today I am glad to watch this video, thanks!

@jokarmaths7771 2 жыл бұрын

amazing ...kzfaq.info/get/bejne/nsqXqdiAt9SpmXU.html

@reidflemingworldstoughestm1394 Жыл бұрын

One of your best videos so far.

@PrithwirajSen-nj6qq Ай бұрын

Enjoyed very much. Waiting for such type of videos. A nice visualised video.

@erikmingjunma9403 6 жыл бұрын

Alternatively: derive the general power mean and explain the intuitions behind them (with the sum fixed, the greater power has more impact when elements are more spread out)

@tsamrawat5448 3 жыл бұрын

Excellent proof dear. Zordaar zabardast zindabaad

@michellegaud4237 2 жыл бұрын

Très belle démonstration géométrique ! Bravo.

@ashleypkumlvu2947 9 ай бұрын

Thank you, I always poor in math, but your lesson truly raises me up. I am in a tremendous excitment of handle some of these difficulties. Thanks again!💕💕💕

@johnchristian5027 6 жыл бұрын

Whenever he said 'mean' I heard 'meme'

@JohnLeePettimoreIII 5 жыл бұрын

Cool explanation. Thanks, amigo!

@rolfdoets 6 жыл бұрын

Very nice demonstration!

@nyujun 5 жыл бұрын

Nice. I am beginning to be addicted to your math problems.

@izakj5094 6 жыл бұрын

Amazing video, please do more of such proofs

@bachirblackers7299 4 жыл бұрын

Hi Mr Presh thanks a lot for this beautiful explanation and believe me nobody can do better than you did . Perfect just perfect .

@AmanKumar-vd1jc 4 жыл бұрын

Gajab Bhai..I heard first time about root mean square

@thecrazypianist8243 5 жыл бұрын

Presh you re just too awesome!!

@yesidlee 4 жыл бұрын

Beautiful demonstration.

@MyOneFiftiethOfADollar 2 жыл бұрын

You, indisputably have the best technical/math presentation platform on the web. I would be filthy rich if I had a buck for every comment you have received begging you to disclose how you pulled this off? The animation and capacity to explain and erase stuff clearly is world class.

@Zack-xz1ph 5 жыл бұрын

I had to learn about the root mean square when I was reading Descartes' Geometry but I never learned this. Fascinating

@prateekgargx 5 жыл бұрын

you can also use concavity of graphs to extend it to infinite positive no.s

@sudheeradakkai5227 4 жыл бұрын

Awesome....thanks...

@aliyardimoglu5629 5 жыл бұрын

Very nice, such a meaningful demonstration..

@udayadityabhattacharyya7496 5 жыл бұрын

Very nice description.

@donaldasayers 6 жыл бұрын

What about Gauss's arithmetic, geometric mean? (Useful for elliptic integrals.)

@andabata43 6 жыл бұрын

Frank K. There is also a rather beautiful generalization: Let t be any real number, and for any POSITIVE x1, x2, ..., xn, define M[t](x1,x2,...,xn) = (Sum[(xk)^t, {k,1,n}])^(1/t). Then if t1 < t2, we have M[t1] ≤ M[t2], with equality iff all xk are equal. In particular, M[-1] = HM, M[0] = GM, M[1] = AM and M[2] = RMS, giving the result in the video. It is also interesting to note that Limit(t -> -Inf) M[t] = min{x1,x2,...,xn} and Limit(t -> +Inf) M[t] = max{x1,x2,...,xn}.

@jampaprasad9339 5 жыл бұрын

Your content is amazing

@Luper1billion 5 жыл бұрын

Thanks, I have to visualize mathematics geometrically to really understand, so this was cool

@twistedsim 6 жыл бұрын

This video was interesting. Thank you

@moonwatcher2001 4 жыл бұрын

Interesting, beautiful and useful. Thanks

@yashvardhan2093 3 жыл бұрын

The RMS is also used in the kinetic theory of gases in thermodynamics

@chellurivenkatasatyanaraya240 3 жыл бұрын

Sir,it is very useful video for all mathematics learner's:-CHVSN as a INDIAN mathematician

@ashokkumarmeher4207 4 жыл бұрын

Nice explanation sir....thank u...

@ieimagine 2 жыл бұрын

Thank-you!

@dlevi67 6 жыл бұрын

+MindYourDecisions I would state the geometric mean - even for two numbers - as ab^(1/2). Differently from RMS, where the use of square and square root would not change with the number of terms, the power (or root) order in a geometric mean will change. I would put the segment at 7:00 at the front and use the fractional power notation for the root: this way it's clear one is always _dividing_ something by the numerosity of the data, then state this for n=2 and only last change the power notation to a root, if you think it's more familiar/easier to understand for people when looking at right triangles. Other than that, nice video and animation; thank you!

@NikhilKumar-im8ls 3 жыл бұрын

A beautiful proof. Thanks

@michaelempeigne3519 6 жыл бұрын

Nice proof, I have never seen such proof although I have known about the inequality

@soumyadeeproy6611 2 жыл бұрын

This video deserves 1M+ likes, bcz it is really super awesome, and super cool idea .. No one ever told me this thing

@ffggddss 6 жыл бұрын

+ Presh: At 3m50s: You can also quickly verify that this altitude is the GM by similar triangles, because a/h = h/b And I really like your geometric demo of that chain of inequalities!! BTW, you might mention that all these means are related by being "functional transforms" of the simple (arithmetic) mean. A "transformed mean," TM, using a monotonic function f, is: TM( ֿx ) = f⁻¹(AM(f( ֿx ))) where ֿx = x[1...n]; AM(f( ֿx )) = (1/n)∑ᵢ₌₁ⁿ f( xᵢ ) So: • when f(x) = x², f⁻¹(x) = √x, and TM = RMS • when f(x) = x, f⁻¹(x) = x, and TM = AM • when f(x) = ln(x), f⁻¹(x) = eˣ, and TM = GM • when f(x) = 1/x, f⁻¹(x) = 1/x, and TM = HM Neat, huh? ;-) PS: I suspect that some property of each function - maybe something involving the second derivative - can be used to arrive at those inequalities, but I haven't delved into that. Actually, looking at the list, I'm getting a very strong hunch . . . Fred

@markusdeserno7321 5 жыл бұрын

Fred: your hunch is correct. This all relies on Jensen’s inequality applied generally to the functions x^a. This leads to the so-called power means, which generalize the four special cases mentioned here.

@mohuyapharikal 4 жыл бұрын

Great..Well done

@nishantrai8830 5 жыл бұрын

That was beautiful bro..

@davidvose2475 3 жыл бұрын

What an elegant set of proofs

@prabirroychowdhury2830 4 жыл бұрын

Excellent.

@anandasilva6986 3 жыл бұрын

thanks for wonderful geometry and you

@bhardwajr01 6 жыл бұрын

I really wanted this video.... Thnx

@nagarjunareddyperam3505 3 жыл бұрын

Our sir taught us He used it in many qns This is a really important and interesting inequality

@bhanupratapkaushal21 5 жыл бұрын

Please make similar type of video on circumcentre, orthocentre, incentre,

@billmarty00001 4 жыл бұрын

Good one. Thanks

@phomthang2621 4 жыл бұрын

thanks you very much.

@Kevincooling1234 6 жыл бұрын

Very beautiful!

@susmitamishra8436 6 жыл бұрын

Thanks very much..... I was able to prove only A.M, G.M and H.M

@gopaldevkota8715 5 жыл бұрын

Thankyou for every things

@gauravbharwan6377 3 жыл бұрын

Bring more like this

@fmakofmako 6 жыл бұрын

Beautiful champ. I liked it and have no criticism.

@daklhs6460 6 жыл бұрын

Beautifull proff.

@atil4 2 жыл бұрын

Thank you!!!!!!

@haradhandatta4824 4 жыл бұрын

Hi,Presh. It's a very nice geometrical proof & showing "equality" by animation. Can we prove that , QM+HM>= AM+GM. Indeed, I cannot. The problem is posted in PASCAL ACADEMY-MATH GROUP.

@francescaamadeiaccardo363 4 жыл бұрын

Brilliant!

@darreljones8645 6 жыл бұрын

In the two-variable case, at least, it's easy to show algebraically equality holds if a=b. Just set the two variables equal, and simplify all four expressions to a.

@aye_djay 6 жыл бұрын

what a great explanation

@icew0lf98 6 жыл бұрын

before you said you made it in desmos, I thought to myself I should make this in desmos lol

@woodchuk1 6 жыл бұрын

How about adding the contraharmonic mean to this? It's always greater than or equal to the RMS for any given data set...essentially it's the arithmetic mean of the squares of all the values divided by the arithmetic mean of the values. So for (3,4) it's equal to 3.571, which is greater than the RMS of 3.536. Could that be interpreted geometrically?

@subhankarpramanik2224 6 жыл бұрын

This is really very helpful.....thnku...sir😁😁😁😁😁

@YamiSuzume 5 жыл бұрын

6:37 That animation seems to use way to much CPU for his PC (volume up)

@najmamuslim1123 5 жыл бұрын

excellent

@brentprim1 2 жыл бұрын

what would a and b have to be in order for the four values to be positive numbers?

@ramachandrannatarajan4234 4 жыл бұрын

Super sum

@ittieh22 4 жыл бұрын

I was so impressed!!!!!!!!!!!

@kshitij7b286 3 жыл бұрын

I solve AM>= GM by intersecting chord theorem let's take line a+b and the draw a circle then we pass another diameter perpendicularly so we have (a+b/2)(a+b/2)=(a+b/2)(a+b/2) by intersecting chord theorem and then I pass chord from the meeting point length of a and b then chord will be x²=ab and x²=ab

@fuminocchi4533 6 жыл бұрын

You're a genius... how you can do such things like that in math...

@mr.coconut2310 6 жыл бұрын

this is the reason i subscribed

@LawatheMEid 6 жыл бұрын

thanks for this proof

@bachirblackers7299 4 жыл бұрын

Hello Mr presh . Hello everyone . When i went further ive found that the intersecting point of HM and RMS ALWAYS LAYS DOWN ON AN EYE SHAPE AND VERY BUTTOM POINT OF THE HM LAYS ON AN OVOID SHAPE ( Yes egg shape not an ellipse neither anoval ) and of course the midpoint of the segment GM LAYS ON AN ELLIPSE .