Cauchy Schwarz Application

[@i_amscarface_the_legend9744]

Basically, we get similar inequalities with more than three numbers, if we use n number the product will be greater than n^2 ! What is nice about that proof is that it is so simple! Thank u Dr.Peyam.

[@MathSolvingChannel]

I have a direct proof. Just expand it, so it is equivalent to prove a/b+a/c+b/a+b/c+c/a+c/b >= 6, then we use the inequality (a1+a2+...+a6)/6 >= (a1*a2*...*a6)^(1/6) Done!

[@TanmayGoswami-b8k]

@@MathSolvingChannel nice one lol

[@toaj868]

What's wonderful about this solution to me is that my first thought upon seeing it was something along the lines of- expanding it out won't take me anywhere but if only I could get away with multiplying only the terms I want to- which is exactly what a dot product would allow one to do!

[@rehaanali8145]

for us AM HM inequality is used but this is a beautiful proof than classics

[@CjqNslXUcM]

Wusste gar nicht, dass du Deutsch sprichst! :) Tolles Video, Grüße aus Dresden!

[@Unchained_Alice]

Haven't heard the name Bunyakovsky for years. Came across him when thinking about a question that could be reduced to the conjecture about certain polynomials generating infinite primes. It was about if all natural numbers can be expressed as the difference of a prime and a square. In fact if his conjecture is true, there are infinite ways to do it!

[@drpeyam]

Interesting!!!

[@Unchained_Alice]

Doesn't work of course for the difference of a square and a prime. For the first, p - x² = N -> x² + N = p But the second case is, x² - p = N -> x² - N = p Clearly the second can be factored for some N values so there are natural numbers that can't be written like that. But x² + N (where N is any natural number) is irreducible over the integers, has a positive leading coefficient, and has gcd of 1 (for x=1,2...) so satisfies the conditions. But the conjecture is unproven sadly. Unrelated to this video but reminded me of it when I heard that name! Pretty interesting at first that the first (probably) always works but the second doesn't always.

[@ahmadtariq3960]

Another nice proof is here . Take g(x)=1/x then g is convex by property of convex function g((x1+x2+x3)/3)

[@drpeyam]

That’s really nice too

[@willyh.r.1216]

I knew Cauchy Schwartz inequality, but I never heard about Bunyakowsky name. Anyway, it's a straightforward elegant proof, thanks 4 posting it Doc.

[@firi4737]

Another proof: Without limiting generality we can say that a, b, c > 0 (if all negative then get two -1 from both factors and get the same with positives). Now use means inequalities for both factors a + b + c >= 3(abc)^1/3. 1/a + 1/b + 1/c >= 3(1/abc)^1/3. Multiply inequalities and get the needed one.

[@archimidis]

Here's a proof using only high school algebra. Since [a+b+c]*[1/a+1/b+1/c]=a/b+b/a+b/c+c/b+c/a+a/c+3 it's enough to show that a/b+b/a+b/c+c/b+c/a+a/c>=6 Due to symmetry it's enough to show that a/b+b/a>=2 or equivalently a^2 + b^2 >= 2ab if we multiply by a*b which is positive by assumption. But this is always true since [a-b]^2>=0 QED

[@the_nuwarrior]

3:04 because x^2 is a monotonic function and keeps the inequality

[@alphaomega3944]

Well done, professor!

[@uelssom]

beautiful proof

[@rikhalder5708]

Very beautiful proof

[@busingyefavour9477]

Maybe apply AM HM inequality to a,b and c

[@Inspirator_AG112]

Do not reduce just cancel the fractions to (1+1+1)!

[@yoav613]

Namely and wlog in the same video,blackpenredpen is so proud of you😃

[@drpeyam]

😂😂😂

[@blackpenredpen]

I am!

[@Happy_Abe]

@@blackpenredpen these comments are amazing!

[@MathSolvingChannel]

I have a direct proof. Just expand it, after subtracting 3 on both sides, then it is equivalent to prove a/b+a/c+b/a+b/c+c/a+c/b >= 6, then we use the inequality (a1+a2+...+a6)/6 >= (a1*a2*...*a6)^(1/6) Done!

[@drpeyam]

It’s nice, but AM GM is as direct as Cauchy schwarz

[@MathSolvingChannel]

@@drpeyam fast reply!

[@evgeniykossiy4649]

You can also notice that in this case you have three pairs of type x+1/x ( a/b+b/a, a/c+c/a, b/c+c/b), and we know that function f(x)=x+1/x has max at (-1,-2)and min at (1,2), therefore if you need same sign a,b and c, you have two options: a,b,c left side 0 -> left side>=6.

[@MathSolvingChannel]

@@evgeniykossiy4649 right, so they are de-composited into 3 pairs, nice!

[@parry9842]

Loved the explanation, Dr. Peyam! Feels great to revisit these beautiful mathematical derivations from when I was preparing for the IITs. The above can be solved directly too, right? I basically expanded the original expression: (a + b + c)(1/a + 1/b + 1/c) = 1 + 1 + 1 + a/b + a/c + b/a + b/c + c/a + c/b. Now, a/b + b/a >= 2*sqrt(a/b * b/a) = 2 Similarly, a/c + c/a >= 2, and b/c + c/b >= 2. The original expression therefore resolves as: (a + b + c)(1/a + 1/b + 1/c) >= 3 + 2 + 2 + 2 = 9.

[@drpeyam]

The sqrt thing is literally Cauchy Schwarz

[@parry9842]

@@drpeyam Agreed, your method was more elegant :-)

[@mohammadrehan8564]

You made it so simple sir❤️

[@MrRyanroberson1]

Consider the 1-dimensional functions a(x) = x, b(x) = 1/x. f = a * b (multiply the outputs) f(1) = 1, in fact f(x) = 1, with the fillable exception of 0. Therefore there is no value of x such that f(x) < 1. Therefore f(x) >= 1. Consider the n+1-dimensional functions a(x,Y) = x + a(Y), b(x,Y) = 1/x + b(Y). b is an associative and commutative function, as is a. f(1,Y) = n+1 when all entries of Y are 1. df(x,Y)/dx = da(x,Y)/dx b(x,Y) + a(x,Y) db(x,Y)/dx da(x,Y)/dx = 1 db(x,Y)/dx = -1/x^2 df(x,Y)/dx = b(x,Y) - a(x,Y)/x^2 = 1/x + b(Y) - 1/x - a(Y)/x^2 = b(Y) - a(Y)/x^2 the derivative of f(x,Y) in x is f(Y). Notice: f(x,Y) = f(-x,-Y); f(0,Y) is undefined. WLOG, assume x > 0. df(1,Y)/dx = b(Y) - a(Y) df(x>1,Y)/dx = b(Y) - a(Y)/x^2 df(00, y=1 is the local minimum. For each x

[@michaelz2270]

I don't think I've seen this proof of the AM-HM inequality before, nice video.

[@antoniocampos9721]

Amazing solution !!

[@tgx3529]

It's fackt, if I use a>0,b>0,c>0 And a+b+c= constant, then I take the method Lagrange of multiplikátor, there are 3" the same "eqv., there is the extrem (minimum) for x=y=z

[@drpeyam]

Works too 😁

[@lazarusisaacng]

I think this proof method called Triangle Inequality, because I found by my school teaching book. Right? And my book has also introduced this proof method called Cauchy-Schwarz Inequality. However, I think AM HM can be proofed easier.

[@dylank6191]

So it should work for a = 2pi, b = 0 and c = -2pi too, shouldn't it? After all, they have the same sine ... :)

[@drpeyam]

No 0 is neither positive nor negative, also 1/0 is not defined

[@dylank6191]

@@drpeyam Yeah I know, I wanted to make a little pun with sign/sine :D

[@chaparral82]

Yes Cauchy-Schwarz works, but in fact it is quite easy. Expand and you will get (a/b +b/a) + (b/c + c/b) + (a/c + c/a) >= 6 replace a/b = p, b/c = q, a/c= r, then you get (p+1/p) + (q + 1/q) + (r + 1/r) >=6. And this is true because p + 1/p >= 2 for any positive p and has a minimum 2 at p = 1.

[@rikhalder5708]

I have a question. What's important to define vector space.Please reply me. 😀😀😀😀😀😃

[@icew0lf98]

let x≥0, then we can prove (1+x + 1/(1+x)) ≥ 2 assume the opposite: 1+x + 1/(1+x) < 2 then x+1/(1+x) < 1 then 1/(1+x) < 1-x multiply both sides by 1+x we get 1 < 1-x^2 which is a contradiction, now from this it follows that a positive number plus its reprocical is greater than 2, now we can just expand (a+b+c)(1/a+1/b+1/c) to get 3+(a/b+b/a)+(b/c+c/b)+(c/a+a/c) ≥ 3+2+2+2=9

[@armacham]

Another way to prove: define f(a,b,c) = (a + b + c)(1/a + 1/b + 1/c) so we are trying to prove f(a,b,c) >= 9 second, establish by parallelism that all positive is the same as all negative, so you only need to consider the case where a,b,c > 0 you can do this by showing f(h, i, j) = f(-h, -i, -j) now multiply it out, you get: a/a + a/b + a/c + b/a + b/b + b/c + c/a + c/b + c/c = 9 3 + (a/b + b/a) + (a/c + c/a) + (b/c + c/b) >= 9 (a/b + b/a) + (a/c + c/a) + (b/c + c/b) >= 6 = 2 + 2 + 2 if you can show prove the following three things, you've got your proof (a/b + b/a) >= 2 (a/c + c/a) >= 2 (b/c + c/b) >= 2 by parallelism, if you prove any one of those three things, you've proven the other two. So you only need to prove that a/b + b/a >= 2 when a and b are both positive real numbers this leads to three cases: ab a=b is easy enough to prove, 2 >= 2, ez by parallelism proving the case where ab so the only thing left to do is to prove that when 0 < a < b, a/b + b/a >= 2; after you prove that, you proved everything let's introduce a new variable z, z > 0 b = a + z substitute that into the equation a/(a+z) + (a+z)/a >= 2 you know that a and a+z are both greater than zero (because a and z are both greater than zero) so you can safely cross-multiply without changing the direction of the "greater than or equal to" sign aa + (a+z)(a+z) >= 2a(a+z) aa + aa + 2az + zz >= 2aa + 2az subtract 2aa from both sides 2az + zz >= 2az subtract 2az from both sides zz >= 0 we know this is true simply because z is a real number, this is true for all real numbers and that's another way to prove the inequality

[@drpeyam]

You don’t need cases for the ab part, just put under a common denominator and use (a-b)^2 >= 0

[@BrynSCat]

sqrt 1 = +- 1 ?

[@drpeyam]

No sqrt(1) = 1

[@BrynSCat]

@@drpeyam I would see it as an event horizon in Projective Geometry Algebra

[@lacuentadevideos]

how about engineers solution to try 1? 1+1+1(1/1+1/1+1/1) = 9 SOLVED!! did I mis something???

[@drpeyam]

Yes you did, you had to show this is true for all a b c

[@lacuentadevideos]

sorry was a probe, not to find a solution

[@BrynSCat]

I think it is only 9 if you ignore the imaginary part,you have probably rotated these using Sqrt ??.

[@drpeyam]

The quantity is positive

[@KarlMarX_93]

Very nice proof! I like C-S inequality. Today I've found very interesting problem. From Fermats Little Theorem we have 2^(p-1)-1 is divided by p. But I can't proof that 2^(p-1) -1 is never divided by p^2. Could you please make video about that? :D Greetings!

[@jameskary5140]

That was super cool

[@mainak8475]

Beautiful

[@tonyhaddad1394]

(a-b)^2 >=0 a^2 -2ab + b^2 >=0 a^2 + b^2 >= 2ab (a^2 + b^2)/ab >= 2 (a/b) + (b/a) >= 2 (wlog suppose a,b,c > 0 ) from here we are done beacaus the expantion of the product is equal to 1+1+1 + (a/b + b/a) + (a/c + c/a) + (b/c + c/b) which is >= 3 + 2 + 2 + 2 = 9 , proof complete !!!! (Negative (a,b,c work the same )

[@drpeyam]

Beautiful

[@tonyhaddad1394]

@@drpeyam thank u 💓💓

[@beautyofmath6821]

Lovely!

[@meinname9747]

I was confused by the german in the end of the video xD

[@drpeyam]

Hahaha

[@Bedoroski]

Classic ❤

[@akshatjangra4167]

Nice!

[@brendanlawlor2214]

Wot a beaudy....never guessed to use the Shwarz inequality 😜

[@azzteke]

beauty - Schwarz!

[@rikhalder5708]

First comment sir. 😀😀😀😄😄

[@emanuellandeholm5657]

9 es gibt doch nichts! :D

[@drpeyam]

Ja genau :D