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Top quality video!

Please use your real voice, AI is just not there yet.

if f(x) is increasing but concave down on some interval including all the xi values you can show pretty easily using the 1st order taylor series with remainder that f(sum (pi*xi)) >= sum (pi *f(xi)). calling that first sum "AM" for convenience, express each xi as AM + ei. note all the remainders are negative where ei is nonzero, and sum (pi*ei) = 0. f(x) = logx completes the original request; f(x) = -1/x proves the HM-GM relationship.

Well that was fun, not sure if I understood it so fast. But I like it. I have never come across this inequality before but then I don't talk to many people. On first sight it seems to offer so much but in fact delivers so little. Maybe that's for why I didn't know about it. My interest though is in that you have a statement about additions in correspondence with multiplications. To have even one is interesting. Maybe there is much more to be learned? This is pretty fundamental to be honest, and totally un understood by masses of Mathematicians. There in is a connection here worthy of a deep dive, but many do not have the breathing capacity :-) Of course I am new to this: add and divide is as to multiply and root. Not many people think like that these days methinks. The inequality is obvious though because the division by n is the same across all x where as the nth root is proportional to each x. It can only get smaller. I realise now that this is statistics duhhhh. I was thinking in terms of Prime Numbers and their mystery. But, and to the point: Anything that relates Addition to Multiplication in the higher indices of numbers is like Gold Dust. Perhaps this is the first Sovereign?

By far the best proof

Bro, please change that voice

Nice proof

An amazing proof I have another one which uses set theory R+^n gives all R+ series which contain n element Define a set that for some s belongs to R+^n Sum of s = SC and Product of s = x We will call that set as P Here s is a seri not number We clearly see that all members of P is smaller than SC^n Therefore there is a smallest reel number that for all x belongs to P x smaller or equal to that reel number We will call that reel number as Pmax Assume that SC = x1+x2...xn Pmax = x1x2...xn xa≠xb SC=x1+x2...(xa+xb)/2...(xa+xb)/2...xn So product of them in P However ((xa+xb)/2)²>xaxb x1x2......(xa+xb)/2...(xa+xb)/2...xn > Pmax A contribiction therefore x1=x2=...=xn After that as easy as pie If there is place you can't get that is bad of my A level english.

Crazy, need some time to digest it, so I got the main point

I knew two proofs of the means inequality. Now I know three; and this is the shortest!

This channel is pure gem

Great problem! Just maybe try to slow down a little, it makes following the video easier

nice

Yeah... definitely not explained in a way an idiot like me would ever understand. Is this the intended effect? You lost me around the 0:40 mark.

www.youtube.com/@MathProblemsGAS

www.youtube.com/@MathProblemsGAS

Ι apply Lagrange multipliers let F(x,ψ ,z,w) =xψzw min(xψzw) =? G(x,ψ,z,w)=1/x^4+1+1/ψ^4+1+1/z^4+1+1/w^4+1 θF/θχ=λθG/θχ Ψzw=λ4χ^3/sqr(x^4+1) θF/θψ=λθG/θψ χzw=λ4ψ^3/sqr(ψ^4+1) by dividing by members we get x=ψ similar ψ=z z=w therefore 4/x^4+1=1 x^4=3 similar ψ^4=z^4=w^4=3 min(xψzw)=3 That is xψzw.>=3 sqr means second power

let f(x) =1/1+x^4 dψ/dx =- 4x^3/sqr(x^4+1) function decreasing a<=b<=c<=d f(d)<=f(c)<=f(b)<=f(a) 4/1+d^4 <=1 d^4>=3 with rearrangement a,b,c,d a^4>=3 b^4>=3 c^4>=3 (abcd)^4>=3^4 abcd>=3 sqr means second power

@ Mathemadix -- You need to address the point that robertveith brought up, if you have not already.

If n is divisible by neither 2 nor 3, then n = 6m +/- 1, for m an integer. Then, n^2 - 1 = (6m +/- 1)^2 - 1 = 36m^2 +/- 12m + 1 - 1 = 36m^2 +/- 12m = 12m(3m +/- 1). Look at two cases: If m is even, then 12m is divisible by 24 and so would n^2 - 1. If m is odd, 3m is also odd. And, that makes 3m +/- 1 even. So, 12(3m +/- 1) would be divisible by 24, and so would n^2 - 1. This covers both cases, and the proof is completed.

*@ Mathemadix -- Your statement on the screen is *not equivalent* to what you read. On the screen "if n is not divisible by 2 & 3" means that n may not be divisible by 2, or n may not be divisible by 3, or n may not be divisible by either 2 or 3. That is different from what you said.

Good question

solved nicely

This channel is pure gem

Golden ratio problem: r=Φ=(1+√5)/2; Plug (Φ^16-1)/(Φ^8+2*Φ^7) into calculator and get 21. If calculator is not allowed then on to Plan B. Students without a calculator should know that Φ^n=F_n*Φ+F_{n-1} Fibonacci says Φ^16-1=987Φ+610-1= 987Φ+609 and Φ^8+2Φ^7=21Φ+13+26Φ+16=47Φ+29 Finally, (987Φ+609)/(47Φ+29) = 21.

this is the only method in my mind that prove AM GM inequality without using mathematical induction

Really nice. Didn't expect it to be explained so simply.

U just blew my mind, thanks 👏👏👏

The solution to the problem at the end : *f(x) = 1-x^2*

Pls do harmonic mean instead

Since i have found your channel, it motivated me to start doing such problems on my own and i want to thank you a lot!

Nice explanation, creative ways to solve them, satisfying animations, and short too. Yet you only have 520 subs?? Well you've got yourself a new One! You deserve it! 🎉

Can you do it with largest exponent function?

Hi Mathemadix, is it in anyway possible to get in touch with you over the email?

Applying the first equation to the second, we get x(x5+x4+...+x+1)=0 and same for y, from those we get x,y=0,-1, then we try the four solutions and get only those where x=y, so (-1,-1) and (0,0). That would be interesting if we took the imaginary numbers too :)

f(x)+f(1/x) = 1 Then f(2x+1)+f(1/2x-1) is equal to....

How to see things like that? Bc it seems to be pulled out of ass for me.

Finally I found a proof for n terms, not just 2 terms.

I didn't understand 2:27

I set all of them equal to each other to find that individually they all had to be a minimum of 3^0.25 since they're symmetric and the reason it is the minimum is because if you shrink any one of them the other three will increase causing the product to increase as well so the case where the product is at it's smallest is when they are equal

Now that's Math content.

simple (a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a)

Very elegant solution...

George Polya was a master in finding elegant proofs . I had the pleasure of meeting him personally on several occasions.

First view and first comment... :nice explanation😅

Holy shit, crazy that such a complicated problem can be solved so easily