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

[@morethejamesx39]

Hey this is Presh Talkwalker

[@JonSebastianF]

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

[@Tehom1]

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

[@ffggddss]

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]

Did you figure this out?

[@darreljones8645]

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

[@Jack_Callcott_AU]

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]

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

[@pe3akpe3et99]

first answer

[@GermansEagle]

Seriously, how have I never heard of this proof.

[@fernandowong5799]

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

[@sunilrampuria9339]

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

[@brodieenrique1003]

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

[@kenzorowen2048]

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

[@abramdrake4510]

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

[@u5s9e2hb4ijk7bv]

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]

how do you know both triangles are congruent?

[@TheOfficialCzex]

Fresh Tall Water. Got it.

[@greg939]

You mean Fresh Saltwater

[@sahilsagwekar]

Presh tall walker

[@sahilsagwekar]

@Sai Sasank presh talwalkar l, his sirname is indian

[@aashsyed1277]

@@sahilsagwekar no he lives in USA

[@mathlegendno12]

@@sahilsagwekar r/whoosh

[@Trinexx42]

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

[@bhardwajr01]

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

[@jaroslavsevcik3421]

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

[@dorijancirkveni]

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

[@marcusyang7686]

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

[@facitenonvictimarum174]

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!!

[@looney1023]

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

[@AnshuKumar-oj8ww]

You have done everything elegantly. Nice immaculate work ! 👍

[@deadfish3789]

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]

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

[@anishkrishnan9698]

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

[@zeynarz7614]

@@anishkrishnan9698 Thanks a lot!

[@parahumour4619]

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

[@koenth2359]

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.

[@shanmugasundaram9688]

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

[@jackthatmonkey8994]

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

[@bernhard5295]

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

[@popogast]

Most useful contribution of the last weeks. Thank You.

[@GermansEagle]

Thats awesome man! Nice video !!!!

[@PhilipBlignaut]

The best description regarding means ever!

[@Etothe2iPi]

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

[@Epoch11]

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]

en.m.wikipedia.org/wiki/Mean

[@raghavagarwal5435]

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

[@jokarmaths7771]

amazing ...kzbin.info/www/bejne/npXKp6iFh9N4n6M

[@iamyoda7917]

Real math! Hooray!

[@titan1235813]

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]

get it, useful in IMO 😂

[@sanseng000]

Simply superb! Super excellent! Awesomely simple.

[@michellegaud4237]

Très belle démonstration géométrique ! Bravo.

[@ahmedbaig7279]

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.

[@notspaso6644]

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

[@erikmingjunma9403]

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)

[@alvarezjulio3800]

What a beauty! That was awesome! Thank Sir!

[@reidflemingworldstoughestm1394]

One of your best videos so far.

[@ashleypkumlvu2947]

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!💕💕💕

[@Zack-xz1ph]

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

[@isaacpark1016]

Beautifully demonstrated. Love it!

[@jokarmaths7771]

amazing ...kzbin.info/www/bejne/npXKp6iFh9N4n6M

[@user-uo8hc1ju4l]

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

[@babitamishra524]

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

[@jokarmaths7771]

amazing ...kzbin.info/www/bejne/npXKp6iFh9N4n6M

[@nyujun]

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

[@MyOneFiftiethOfADollar]

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.

[@johnchristian5027]

Whenever he said 'mean' I heard 'meme'

[@prateekgargx]

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

[@izakj5094]

Amazing video, please do more of such proofs

[@AmanKumar-vd1jc]

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

[@PrithwirajSen-nj6qq]

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

[@ankitjain3760]

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.

[@bachirblackers7299]

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

[@ffggddss]

+ 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]

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.

[@donaldasayers]

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

[@yesidlee]

Beautiful demonstration.

[@rolfdoets]

Very nice demonstration!

[@JohnLeePettimoreIII]

Cool explanation. Thanks, amigo!

[@yashvardhan2093]

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

[@andabata43]

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}.

[@aliyardimoglu5629]

Very nice, such a meaningful demonstration..

[@tsamrawat5448]

Excellent proof dear. Zordaar zabardast zindabaad

[@dlevi67]

+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!

[@davidvose2475]

What an elegant set of proofs

[@turtlellamacow]

Finally a respectable video from this channel! How have I never seen this geometric argument

[@soumyadeeproy6611]

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

[@thecrazypianist8243]

Presh you re just too awesome!!

[@bhanupratapkaushal21]

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

[@michaelempeigne3519]

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

[@NikhilKumar-im8ls]

A beautiful proof. Thanks

[@Luper1billion]

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

[@chellurivenkatasatyanaraya240]

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

[@haradhandatta4824]

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.

[@nagarjunareddyperam3505]

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

[@udayadityabhattacharyya7496]

Very nice description.

[@moonwatcher2001]

Interesting, beautiful and useful. Thanks

[@YamiSuzume]

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

[@twistedsim]

This video was interesting. Thank you

[@woodchuk1]

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?

[@icew0lf98]

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

[@jampaprasad9339]

Your content is amazing

[@darreljones8645]

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.

[@fuminocchi4533]

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

[@ashokkumarmeher4207]

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

[@calyodelphi124]

Something I would like to point out is that a slight variation of the harmonic mean is also used in electrical engineering for parallel resistances/inductances and series capacitances. I call it the parallel mean, but it's also just referred to as "x parallel y parallel ...": HM(x1,x2,...,xn)/n = 1/( 1/x1 + 1/x2 + ... + 1/xn ) = (sum(i = 0 to n) xi^-1)^-1 It's often written in EE as x1||x2||...||xn, and it's calculated exactly as the harmonic mean would be, except that instead of dividing n by the sum of the reciprocals of the values, you instead take the reciprocal of the sum of the reciprocals, since multiple parallel resistances/inductances or series capacitances act as if they are a single equivalent series component in the circuit.

[@franzluggin398]

Well, that's just the same as the harmonic mean, you just tack on a division by n at the end. That's not really enough difference to justify talking about it in this video. It's also not, strictly speaking, an average, or mean, since the parallel resistance of A and B is not going to be between A and B except in contrived edge cases (A or B = 0 are the only ones I can think of right now). But thinking of your scenario in terms of the harmonic mean is probably the best way to see the difference: The way I think about it is that the harmonic mean accurately describes the correct average to take over the parallel resistances (and similar parts). If you calculate HM(R1, R2, R3, ..., Rn), then you get the resistance that some hypothetical part would need to have so that you could exchange every resistance R1, ..., Rn with that one part (so all resistances are the same) without altering the total resistance in this circuit. But since you're not actually interested in the mean, but the total resistance, you divide the number by n, since all paths now have equal resistance, so having n paths will lower the resistance to 1/n-th of that of one part.

[@calyodelphi124]

@Franz: That is actually a really good way to think about it, and not a way to think about it that I'd thought about before! I think when I finally get off my keister to do a video about parallel resistances, I'll have to use this, because this is a really good description of how it works! :)

[@fmakofmako]

Beautiful champ. I liked it and have no criticism.

[@sudheeradakkai5227]

Awesome....thanks...

[@susmitamishra8436]

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

[@nishantrai8830]

That was beautiful bro..

[@pholioschenouda5395]

What program do ypu use to illustrate your problems???

[@markgraham2312]

What about the median and the mode? They, too, are also types of arithmetic means.

[@fizixx]

Very interesting! This is one of my favorites! Thanks

[@mohuyapharikal]

Great..Well done

[@bachirblackers7299]

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 .

[@bhardwajr01]

I really wanted this video.... Thnx

[@brentprim1]

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

[@shivenrathore805]

i have the algebric proof without using contradiction and I knew this inequalities a long before andthe inequalities only work if both a and b are positive

[@harshitgarg6483]

archana rathore works also for negative but the signs flip

[@srinathdas398]

Sir,, if we know the diference of 'a' and 'b', and there is a problem like R.M.S. = x(A.M.), can we define 'x'??

[@gauravbharwan6377]

Bring more like this

[@anandasilva6986]

thanks for wonderful geometry and you

[@prabirroychowdhury2830]

Excellent.

[@ieimagine]

Thank-you!

[@manla8397]

This is a beautiful proof.

[@dreamchaser9272]

Sir pls tell us similar channels like yours for physics. I am preparing for physics olympiad

[@kshitij7b286]

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

[@subhankarpramanik2224]

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

[@philipcollier4883]

Nice video. I don't know the answer but is there a value of a and b where exactly 2 of the means are equal to eachother but the rest are not? How about for the mean of N terms?

[@JannisAdmek]

Philip Collier no, as far as I can tell

[@daklhs6460]

Beautifull proff.

[@gopaldevkota8715]

Thankyou for every things

[@xaxuser5033]

Nice geometric proof i knew just how to prove it by algebra