Quick method : AM - HM inequality ( x + y + z ) / 3 > or = 3 / [ ( 1 / x ) + ( 1 / y ) + ( 1 / z ) ] (x + y + z ) * [ ( 1 / x ) + ( 1 / y ) + ( 1 / z ) ] > or = 9
This a straight forward application of the AM-GM inequality! There was a saying we say a lot in math Olympiad "All inequalities can be proven using the AM-GM inequality in a suitable manner" 😁
Application of AM-GM inequality gives us (x+y+z) >= 3 × cube root of xyz and (1/x+1/y+1/z) >= 3 × cube root of 1/xyz. Multiply the inequalities to get the reqd result. (Note that x, y and z are positive and hence we could use the inequality.)
We can indeed : Assuming all x_i are positive Let < | > be the canonical inner product in Rⁿ and || || its euclidian norm. The Cauchy-Schwarz inequality in Rⁿ with X:=(√x_i)_{1 ≤ i ≤ n) and Y:=(1/√x_i)_{1 ≤ i ≤ n} yields : ||X|| ||Y|| ≥ || i.e (by squaring both sides) : ||X||² ||Y||² ≥ ² Which is equivalent to : Σ((√x_i)², i=1, n) * Σ((1/√x_i)², i=1, n) ≥ [Σ(√x_i*1/√x_i, i=1, n)]² Since all x_i are positive and Σ(√x_i*1/√x_i, i=1, n)=Σ(1, i=1, n) = n, we have : Σ(x_i, i=1, n) * Σ(1/x_i, i=1, n) ≥ n² And as a special case (for n = 4), we get that : For any x, y, z, t > 0, (x+y+z+t)(1/x + 1/y + 1/z + 1/t) ≥ 4² = 16
Symmetry x=y=z == 1 is the minimum as (1+ 1 + 1)(1+1+1) >= 9 at 9=9 if you perturb any variable the inequality increases. A.1/A = 1 so a 9 can be found 3.Ax 3.1/A == 9.1 :)
@@thangnguyen-iw8tb just looked it up on wikipedia real quick, and in fact sometimes it's even called Cauchy - Bunyakovsky - Schwarz inequality. And yea, it's odd to hear another name apart from the usual two
We can define F(x,y,z) = (x+y+z)(1/x + 1/y + 1/z) and use multivariable calculus to find the minimum. When we do this, we find the minimum occurs when x² = yz, y² = xz and z² = xy. _These equations tell us that_ _xyz = x³ = y³ = z³_ _So, x = y = z, when F(x,y,z) is a minimum, as x, y and z are all positive._ _So the minimum of_ _F(x,y,z)_ _= (x + x + x)(1/× + 1/x + 1/x)_ _= (3x)(3/x) = 9_ _So, F(x,y,z) ≥ 9, but let's go for overkill 😁_ If we substitute x² = yz into F(x,y,z), we find F(x,y,z) = (1 + y/x + z/x)² after several algebraic step = (1 + y/x + x/y)², as x² = yz, so x/y = z/x So, we need to show that (1 + y/x + x/y) ≥ 3. We can now define G(x,y) = 1 + y/x + x/y and using multivariable calculus (again), we find the minimum occurs when x = y. So, (1 + y/x + x/y) = 3 is the minimum of G(x,y), as x = y. So, F(x,y,z) ≥ (1 + y/x + x/y)² = 3² = 9. Long story short. Multivarable calculus can solve this problem, but using inequalities is way easier and aesthetically more pleasing 😁
I blundered around with this for a bit and came to within a few yards of your #1 result, but got lost in the weeds! Thanks for taking me by the hand and leading me the rest of the way!
Ok before watching, I did it like this: Held y and z constant, solved for x where the expression is at a minimum (using calculus): x = sqrt(yz) Sub this in, hold z constant and repeat for 'y': y = z Thus, it can be found that x=sqrt(y^2)=sqrt(z^2), that is: x=y=z. Sub this back in and we find that: min(function) = (x+x+x)(1/x+1/x+1/x) = 3x*3/x = 9.
For the summation 2+3+6+7+10+11+14+15+.............+(4x+2)+(4x+3)=Z^n, where n>1, how many +ve integral solutions for Z are possible? Dare to solve this simple-looking, Elegant & tough problem?
Another nice solution is to evaluate the minimum of f(x,y,z)= (x + y + z)*(1/x + 1/y + 1/z) by solving df = 0 . One gets x = y = z = 1 with min(f(x,y,z)) = 9. This also works for more variables (f(x1, x2, .... xn)
@@SyberMath What we need is the gradient of f. Each component of the gradient must be zero. Accordingly after some manipulations we get three equations: x = yz; y = xz and z = xy. If we solve that, we get one positive solution: x = y = z = 1.
O.k., what then still is to be done is to show that that is a minimum, but for the symmetrie of the problem this will be done by evaluating the function just anywhere.
You can also use Sedrakyan's inequality to prove that 1/x+1/y+1/z is always greater or equal to 9/(x+y+z) 1/x+1/y+1/z >= ((1+1+1)^2)/(x+y+z)=9/(x+y+z) for any positive reals x,y,z Therefore, (x+y+z)(1/x+1/y+1/z)>=(x+y+z)*9/(x+y+z)=9 for any positive reals x,y,z The equality is only attained for x=y=z
Quite complicated, just distribute it and then replace x/y = u, y/z = v and z/x = w. Then you will receive u+1/u + v+1/v + w + 1/w >= 6 It is easy to show that x + 1/x >= 2 if x is positive. q.e.d
(x+y+z)(1/x+1/y+1/z)=3+([u+1/u]+[v+1/v]+[w+1/w]) where u=x/y,v=y/z and w=z/x. now u+1/u=(u^2+1)/u>=2 ...etc. because (u-1)^2>=0. Hence the above RHS in the equation is >=9. Hence the result.
I liked the 2nd method as usual, bcoz the first method doesn't seems to be obvious to me, líke thinking of 2 inequality, multiplying them, replacing x with xy, the second method was short and good too 😁
Uhh of course I do,but bprp and Dr peyam call It the Chen Lu and so do their fans too. Besides be glad that I read your comment,besides this is not how youd reply to someone, but your 9 so its kind of ok,when youll age youll reply properly I think.
I won't get tired of seeking what I have no knowledge about. I'm always getting new subscribers yet getting fewer views. And I've applied different logics 😔
Hmm yes the floor is made out of floor. Memes aside,Your awesome Syber so keep It up and I found away to do integrals (where Partial Fractions or pf the short cut is intended) without pf.
Hello sybermath this is an interesting problem Btw are you sick? Seems like you have a cold? I hope not but if you have a cold I hope you get better soon
I prefer the second method it’s very soft and cute. Towards the first one wich is less nice 😎. But can we talk about cute mathematics ? I think yes. when most people talk about mathematics as an ugly studies.
solved it in my head while hanging the laundry... (x+y+z)(1/x+1/y+1/z) >= 9 so let's multiply all that: x/x +x/y+x/z+y/x+y/y+y/z +z/x+z/y+z/z >= ?? now simplify and reorganize: 1+ x/y +y/x + 1 + y/z + z/y + 1 +z/x +x/z >=?? 3 + (x/y + y/x) + (y/z + z/y) + (z/x + x/z) >= ?? not let's take closer look at (x/y +y/x) and try to calculate how much it is and we can assume same reasoning can be applied to other 2 similiar parts of out equation: x/y + y/x = ? let's make it one fraction: x*x/yx + y*y/yx = (x^2 + y^2)/xy let's assume (x^2+y^2)/xy = k x^2 +y^2 = kxy x^2 - kxy +y^2 =0 Let's solve the above for x. We count delta = b^2 -4ac, and our a= 1 b= -ky c= y^2 we get delta = (-ky)^2 -4y^2 delta = k^2y^2-4y^2 = y^2(k^2-4) for equations to have a real solution delta needs to be >=0 so y^2(k^2-4) >=0 and y^2 cannot be 0, because y >0 k^2-4 >=0 k^2 >= 4 k >=2 or k =2 or x/y+y/x =2 and that leads back to our equation: 3 + (x/y + y/x) + (y/z + z/y) + (z/x + x/z) and since all 3 parts in parenthesis are similiar (we can solve each of them in the same way as x/y +y/x) they all must be greater or equal 2 which means, our whole equation is greater or equal to 9
Ingilizce sonra da Turkce kafa karistirici olur izleyiciler icin. Ingilizce ogrenirsen anlayabilirsin. Hem bu senin icin cok iyi bir sey olur. Hem de ortami kolaylastirmak yerine ortama kendini uydurmus olursun.
@@SyberMath Siz gordunuz mu sorulari bilmiyorum ama bence zordu pek cok baska sinava kiyasla ya ben 34 isaretledim kodlama hatam yoksa dogru hepsi ama pek cok kisi (en azindan cevremdeki) 30 zor isaretledik diyor umarim hakkimizda hayirlisi olur
Bu arada vesile oldu tekrardan bu kanali actiginiz icin cooook tesekkurler pek cok insan icin muthis bir is gerceklestiriyorsunuz gerci benim de biraz kafami celmiyor degilsiniz turkiyede yasadigimi unutup matematik okusam mi diye ama :D anyway tekrardan sonsuz minnet ❤❤
Hello everyone! Here is the link to Math Elite's channel! Check it out:
ruclips.net/user/mathelitefeatured
I subscribe this channel.
Quick method :
AM - HM inequality
( x + y + z ) / 3 > or = 3 / [ ( 1 / x ) + ( 1 / y ) + ( 1 / z ) ]
(x + y + z ) * [ ( 1 / x ) + ( 1 / y ) + ( 1 / z ) ] > or = 9
Now all you have to do is prove the AM-HM inequality 😂🤣.
@@davidbrisbane7206 that's easy haha
@@diogenissiganos5036
🤣😂🤣😂
@@davidbrisbane7206 there is a geometric proof which is worth looking into
@@diogenissiganos5036
Indeed there is geometric proof, but it only for two variables I think.
am-hm proves this is 10 seconds
also with cauchy-schwartz it is pretty easy
yes
This a straight forward application of the AM-GM inequality!
There was a saying we say a lot in math Olympiad "All inequalities can be proven using the AM-GM inequality in a suitable manner" 😁
😁
Application of AM-GM inequality gives us (x+y+z) >= 3 × cube root of xyz and (1/x+1/y+1/z) >= 3 × cube root of 1/xyz. Multiply the inequalities to get the reqd result. (Note that x, y and z are positive and hence we could use the inequality.)
“Cauchy-Schwarz inequality”
I know that one from Bri on one of his videos.
It looks pretty awesome.
14:34 Can we extend it for 4 values x, y, z, t or even find a formula for n values : (x1 + x2 + … + xn)(1/x1 + 1/x2 + … + 1/xn) > X ?
We can indeed :
Assuming all x_i are positive
Let < | > be the canonical inner product in Rⁿ and || || its euclidian norm.
The Cauchy-Schwarz inequality in Rⁿ with X:=(√x_i)_{1 ≤ i ≤ n) and Y:=(1/√x_i)_{1 ≤ i ≤ n} yields :
||X|| ||Y|| ≥ ||
i.e (by squaring both sides) :
||X||² ||Y||² ≥ ²
Which is equivalent to :
Σ((√x_i)², i=1, n) * Σ((1/√x_i)², i=1, n) ≥ [Σ(√x_i*1/√x_i, i=1, n)]²
Since all x_i are positive and Σ(√x_i*1/√x_i, i=1, n)=Σ(1, i=1, n) = n, we have :
Σ(x_i, i=1, n) * Σ(1/x_i, i=1, n) ≥ n²
And as a special case (for n = 4), we get that :
For any x, y, z, t > 0,
(x+y+z+t)(1/x + 1/y + 1/z + 1/t) ≥ 4² = 16
@@juauke Thanks!
@@goodplacetostop2973 you're welcome ;)
Very nice!
N^2
It was fun. I wrote lol in my notes looking at the long method..
AM>=HM will solve in a couple of seconds though..
Symmetry x=y=z == 1 is the minimum as (1+ 1 + 1)(1+1+1) >= 9 at 9=9 if you perturb any variable the inequality increases. A.1/A = 1 so a 9 can be found 3.Ax 3.1/A == 9.1 :)
well not realy, this function gives 9 anytime x=y=z, not only when they are equal to 1 ;)
The Cauchy - Schwartz inequality would be best here imo
AM ≥ HM will work best here..
@@ayush7800 Thinking about it, they're both straightforward in the same way
in Vietnam they call it Bunyakovsky,weird right?
@@thangnguyen-iw8tb stated by Cauchy proven by bunyakovski and integral form by Herman Schwarz i guess
@@thangnguyen-iw8tb just looked it up on wikipedia real quick, and in fact sometimes it's even called Cauchy - Bunyakovsky - Schwarz inequality. And yea, it's odd to hear another name apart from the usual two
We can define
F(x,y,z) = (x+y+z)(1/x + 1/y + 1/z) and use multivariable calculus to find the minimum.
When we do this, we find the minimum occurs when
x² = yz, y² = xz and z² = xy.
_These equations tell us that_
_xyz = x³ = y³ = z³_
_So, x = y = z, when F(x,y,z) is a minimum, as x, y and z are all positive._
_So the minimum of_
_F(x,y,z)_
_= (x + x + x)(1/× + 1/x + 1/x)_
_= (3x)(3/x) = 9_
_So, F(x,y,z) ≥ 9, but let's go for overkill 😁_
If we substitute x² = yz into F(x,y,z), we find
F(x,y,z)
= (1 + y/x + z/x)² after several algebraic step
= (1 + y/x + x/y)², as x² = yz, so x/y = z/x
So, we need to show that (1 + y/x + x/y) ≥ 3.
We can now define G(x,y) = 1 + y/x + x/y and using multivariable calculus (again), we find the minimum occurs when x = y.
So, (1 + y/x + x/y) = 3 is the minimum of
G(x,y), as x = y.
So, F(x,y,z) ≥ (1 + y/x + x/y)² = 3² = 9.
Long story short. Multivarable calculus can solve this problem, but using inequalities is way easier and aesthetically more pleasing 😁
Replacing x with xy expresses the elegance of the solution.
Oooh , Very Interesting . Heart filled with joy
💖
Damn it, I remember getting this task on local olympiad and failing … :(
Very nice. I was approaching the 2nd method but didn't figure it out.
Glad it helped!
I blundered around with this for a bit and came to within a few yards of your #1 result, but got lost in the weeds! Thanks for taking me by the hand and leading me the rest of the way!
Absolutely!
Please solve this question next!
4abc=(a+3)(b+3)(c+3) where a,b,c are integers which are all greater than all equal to 4
Assume a ≥ b ≥ c,
or that a + 3 ≥ b + 3 ≥ c + 3, and you'll make progress.
I noticed that your second method is alwaus nicer than the first method, to make us watch the all video😀
Thanks! 😃
Ok before watching, I did it like this:
Held y and z constant, solved for x where the expression is at a minimum (using calculus):
x = sqrt(yz)
Sub this in, hold z constant and repeat for 'y':
y = z
Thus, it can be found that x=sqrt(y^2)=sqrt(z^2), that is: x=y=z.
Sub this back in and we find that:
min(function) = (x+x+x)(1/x+1/x+1/x) = 3x*3/x = 9.
Nice!
Best channel i have found so far, can we do beautiful questions of permutations and combinations
Yes we can
Was an amazing session as always!!
Thank you! 😊
Very nice. I like both methods.
Thank you! 😊
For the summation 2+3+6+7+10+11+14+15+.............+(4x+2)+(4x+3)=Z^n, where n>1, how many +ve integral solutions for Z are possible? Dare to solve this simple-looking, Elegant & tough problem?
Your channel is so useful and fun.
Happy to hear that!
Another nice solution is to evaluate the minimum of f(x,y,z)= (x + y + z)*(1/x + 1/y + 1/z) by solving df = 0 . One gets x = y = z = 1 with min(f(x,y,z)) = 9.
This also works for more variables (f(x1, x2, .... xn)
How do you evaluate df?
@@SyberMath What we need is the gradient of f. Each component of the gradient must be zero. Accordingly after some manipulations we get three equations: x = yz; y = xz and z = xy. If we solve that, we get one positive solution: x = y = z = 1.
O.k., what then still is to be done is to show that that is a minimum, but for the symmetrie of the problem this will be done by evaluating the function just anywhere.
Thanos: I used the inequality to prove the inequality
You can also use Sedrakyan's inequality to prove that 1/x+1/y+1/z is always greater or equal to 9/(x+y+z)
1/x+1/y+1/z >= ((1+1+1)^2)/(x+y+z)=9/(x+y+z) for any positive reals x,y,z
Therefore, (x+y+z)(1/x+1/y+1/z)>=(x+y+z)*9/(x+y+z)=9 for any positive reals x,y,z
The equality is only attained for x=y=z
In my opinion the 2 methode are cool , 1st one help us to solve a more complicated inequality the 2nd one is less pain !!!
Quite complicated, just distribute it and then replace x/y = u, y/z = v and z/x = w.
Then you will receive u+1/u + v+1/v + w + 1/w >= 6
It is easy to show that x + 1/x >= 2 if x is positive. q.e.d
Am gm inequality
(x+y+z)(1/x+1/y+1/z)=3+([u+1/u]+[v+1/v]+[w+1/w]) where u=x/y,v=y/z and w=z/x.
now u+1/u=(u^2+1)/u>=2 ...etc. because (u-1)^2>=0.
Hence the above RHS in the equation is >=9. Hence the result.
By Cauchy-Schwarz,
[(√x)^2+ (√y)^2+ (√z)^2)][(1/(√x))^2 + (1/(√y))^2 + (1/(√z))^2)]
= (x+y+z)(1/x+1/y+1/z)
≥ (1+1+1)^2
= 9.
Awesome !!! as always good job !!!
Thank you so much 😀
A math problem a day keeps the online classes away 🤣🤣🤣🤣🤣🤣🤣🤣
Absolutely! 😂
You can solve it in just 5s by using cauchy-schwarz inequality or AM-HM
Another great explanation, SyberMath!
Glad you think so, Carlos! 😊
Good idea
Thanks 😊😊😊😊😊👍👌👍☺️great explanation 😊😄😊😄
Glad you liked it
What is condition of x,y,z so that the inequality equal to 9?
x=y=z
we can use Cauchy Schwarz inequality right?
I think so
Oooh this sounds very interesting
protecting my comment from "don't be evil" but evil youtube spam detectors
@@MathElite compi says 𝚃𝚛𝚞𝚎
@@MathElite are you a member?
Crossover video when? 😛
@@goodplacetostop2973 hello ....
Wow syber you are awesome good example
Thank you!
I liked the 2nd method as usual, bcoz the first method doesn't seems to be obvious to me, líke thinking of 2 inequality, multiplying them, replacing x with xy, the second method was short and good too 😁
Excellent!
i first thought of cauchy-schwartz inequality
Now me ready by chen lu do you mean chain rule?
Uhh of course I do,but bprp and Dr peyam call It the Chen Lu and so do their fans too.
Besides be glad that I read your comment,besides this is not how youd reply to someone, but your 9 so its kind of ok,when youll age youll reply properly I think.
@@3r3nite98 9 years old I am.
Clever. Like the first method though.
I won't get tired of seeking what I have no knowledge about.
I'm always getting new subscribers yet getting fewer views. And I've applied different logics 😔
That's weird
@@SyberMath 😔😔
Using CBS we get (x+y+z)(1/x+1/y+1/z)>=(1+1+1)^2=9 =)))
can you please prove the following:
let :a+b+c = 1 and a,b,c are non-negative
prove that 1/(1-a*a) + 1/(1-b*b) + 1/(1-c*c) > 189/62
Just do AM>=HM
Hmm yes the floor is made out of floor.
Memes aside,Your awesome Syber so keep It up and I found away to do integrals (where Partial Fractions or pf the short cut is intended) without pf.
Hello sybermath this is an interesting problem
Btw are you sick? Seems like you have a cold? I hope not but if you have a cold I hope you get better soon
I'm not feeling well. Thanks for asking. 💖
That's why I sounded that way
Oh I hope you get better soon
I prefer the second method it’s very soft and cute. Towards the first one wich is less nice 😎. But can we talk about cute mathematics ? I think yes. when most people talk about mathematics as an ugly studies.
The famous mathematician G. H. Hardy said: "Beauty is the first test: There is no permanent place in the world for ugly mathematics."
Wonderfull
Thanks
Veeeery Advanced however super interesting
Thank you
Nyc vedio
You should make a discord sever I would love that
Thank you for the suggestion! I'm planning to. Hopefully when I have some time...
Cosy
Shorts
@@MrDowntownjbrown 😅
solved it in my head while hanging the laundry...
(x+y+z)(1/x+1/y+1/z) >= 9
so let's multiply all that:
x/x +x/y+x/z+y/x+y/y+y/z +z/x+z/y+z/z >= ??
now simplify and reorganize:
1+ x/y +y/x + 1 + y/z + z/y + 1 +z/x +x/z >=??
3 + (x/y + y/x) + (y/z + z/y) + (z/x + x/z) >= ??
not let's take closer look at (x/y +y/x) and try to calculate how much it is and we can assume same reasoning can be applied to other 2 similiar parts of out equation:
x/y + y/x = ?
let's make it one fraction:
x*x/yx + y*y/yx = (x^2 + y^2)/xy
let's assume
(x^2+y^2)/xy = k
x^2 +y^2 = kxy
x^2 - kxy +y^2 =0
Let's solve the above for x.
We count delta = b^2 -4ac, and our
a= 1
b= -ky
c= y^2
we get delta = (-ky)^2 -4y^2
delta = k^2y^2-4y^2 = y^2(k^2-4)
for equations to have a real solution delta needs to be >=0
so
y^2(k^2-4) >=0
and y^2 cannot be 0, because y >0
k^2-4 >=0
k^2 >= 4
k >=2 or k =2 or x/y+y/x =2
and that leads back to our equation:
3 + (x/y + y/x) + (y/z + z/y) + (z/x + x/z)
and since all 3 parts in parenthesis are similiar (we can solve each of them in the same way as x/y +y/x) they all must be greater or equal 2
which means, our whole equation is greater or equal to 9
Couldn't you just do the AM-GM inequality?
Is basically the second method
Even if x y z are all negative, this inequality still works
If one is negative, others are positive it does not work
Hocam,eski videolar gibi türkçe videolar da gelse az da olsa ,böyle anlamak zor oluyor:/
Ingilizce sonra da Turkce kafa karistirici olur izleyiciler icin. Ingilizce ogrenirsen anlayabilirsin. Hem bu senin icin cok iyi bir sey olur. Hem de ortami kolaylastirmak yerine ortama kendini uydurmus olursun.
Peki ,hocam teşekkürler.Bunun için önerebileceğiniz yabancı kaynak var mı ,bu tarz sorular için veya konu anlatımı (Cebir için ya da Analiz?
@SyberMath Have you checked my problems
Where did you submit them?
The link that you left(the forms one)
Bài bất đẳng thức quá đơn giản
Magic methode is AM-GM
I liked the 2nd method better.
Quick method is a better proof
All jee students please give attendence!!
😁😊👍
Abi ayt 2021 sorularını çözsen hiç yadırgamam :D
Sorularin telif hakkini bir sirket satin almis ve youtube'da cozenlere dava aciyor diye duydum. Eee kafadan para 😂
@@SyberMath Siz gordunuz mu sorulari bilmiyorum ama bence zordu pek cok baska sinava kiyasla ya ben 34 isaretledim kodlama hatam yoksa dogru hepsi ama pek cok kisi (en azindan cevremdeki) 30 zor isaretledik diyor umarim hakkimizda hayirlisi olur
Bu arada vesile oldu tekrardan bu kanali actiginiz icin cooook tesekkurler pek cok insan icin muthis bir is gerceklestiriyorsunuz gerci benim de biraz kafami celmiyor degilsiniz turkiyede yasadigimi unutup matematik okusam mi diye ama :D anyway tekrardan sonsuz minnet ❤❤
me who knows C-S
ez
Cái này ez