As x is strictly positive, we can multiply both sides by x, giving us that x^4 +1 >=2x^2, form this subtract 2x^2 form both sides leaving us with x^4-2x^2+1>= 0 this can be simplified into (x^2-1)^2 >=0 this is always true as x^2 is always greater then zero
For the first provlem we can transform it to a + b^2 + b/a - 3b >= 0 We can multiply it by a since a > 0 a^2 + ab^2 + b - 3ab >= 0 Now I want to transform some part of left hand side into a quadratic form a^2 + ab^2 + b^2 - b^2 + b - 2ab - ab >=0 a^2 - 2ab + b^2 + ab^2 - b^2 + b - ab >= 0 (a-b)^2 + b^2(a-1) + b(1-a) >= 0 Now I notice that (a-b)^2 is always non-negative, therefore we have to prove that the remaining part is non-negative b(1-a) is the same as -b(a-1) which means that we have to prove that b^2(a-1) - b(a-1) >= 0 b^2(a-1) >= b(a-1) But here I have the issue because for 0
@user-ee4tz1zf4o your are close but not quite there. Even though what you have done is correct, the path you took does not lead you to the answer, you have to think of something else. A direct answer would be to apply AM-GM like mentioned in the video.
Here is my attempt at solving the last problem: Multiply both sides by x so that we have: x^4+1>=2x^2 Now move x^2 to the left side such that we get: x^4-2x^2+1>=0 Split the middle term into two terms with coefficients that sum to -2 and multiply to +1: x^4-x^2-x^2+1>=0 Factor by grouping: x^2(x^2-1)-(x^2-1)>=0 Factor out (x^2-1): (x^2-1)(x^2-1)>=0 Simplify: (x^2-1)^2>=0 Since any real number squared is greater than or equal to zero, this inequality must be true for all x on the real domain. Please let me know if I made a mistake, as it is very likely that I did; I don't have very much experience with proofs and would love to learn.
You did it wrong. When dealing with these kinds of inequalities you should check for when x is negative. You basically completely cross out that possibility. It should look something like this : (x^4+1)/x>=2x^2,(x^4-2x^2+1)/x>=0. From where you get all real positive numbers with the interval method. I apologize for my bad English
My idea of the solution. a + b² + b/a ≥ 3b can be transformed to ab² + (1-3a)b + a² ≥ 0 a>0 and b>0, so ab² > 0. Also, a²>0. If (1-3a)b would be ≥ 0 also, then the sum ab² + (1-3a)b + a² would be ≥ 0. But (1-3a)b≥0 only if a≤1/3. So I decided to determine under what values of _a_ this polynomial is ≥ 0. Consider ab² + (1-3a)b + a² as a quadratic polynomial with respect to b. If the polynomial is ≥ 0, then it has 0 or 1 root (not 2), therefore its discriminant is ≤ 0. D = (3a-1)²-4a³ D = -4a³ + 9a² - 6a + 1 So, if ab² + (1-3a)b + a² ≥ 0, then D≤0 -4a³ + 9a² - 6a + 1 ≤ 0 The usual strat for solving 3rd degree polynomial equations is to guess one of the root, which, in this case, is conveniently appeared to be a=1. Then use long polynomial division to factorize that. -(a-1)²(4a-1) ≤ 0 a∈[0.25;+∞) So. If a∈(0; 0.25), then ab², (1-3a)b and a² are positive, therefore their sum ab² + (1-3a)b + a² is positive. If a∈[0.25;+∞), then ab² + (1-3a)b + a² has 0 or 1 roots, therefore ab² + (1-3a)b + a² is positive for all values of a,b or negative for all values of a,b. You can check that by randomly testing one of its values. For example, if a=2 and b=2, ab² + (1-3a)b + a² is 2. Therefore it's positive ∀b>0 and a≥0.25.
For the final problem: Rearrange (since positive) x^4+1≥2x^2, now rearrange again, and we have x^4-2x^2+1≥0, but this is trivial since (x^2-1)^2≥0 for all x, equality holds when x^2=1, hence x=1
(x^2-1)^2>=0 x^4-2x^2+1>=0 x^4+1>=2x^2 Since x is positive dividing both sides by x doesn’t change inequality. (x^4+1)/x >=2x and equality happens when x=1
When I looked at this my first thought was a proof by induction. Start by noting a isn’t 0. Then if we take b=0 as the base case I can see conceptually the induction hypothesis of b=n+1 could be messy, but I think it would still work.
x^3 + (1/x) is not equal to (x^3 + 1) so your solution is not true. Multiply both sides by x (x is not equal to 0) and you get x^3 -2x +1 >=0, you can get the roots of the equation by using rufinni with x=1 and the second grade equation formulla. You get that that equation is negative between the interval of 1 and 1.618, you can verify that if you put x=1.1 on the original question, it is not true
Since (x^2-1)^2 >=0 Because square of any real number gives a positive value. So, x^4 - 2x^2 + 1 >= 0 x^4 + 1 >= 2x^2 Now since x>0 so, (x^4 + 1)/x >= 2x At x=1 LHS=RHS
the solution feels a little random though it is probably what you are expected to use in a math test especially considering that AM-GM inequalities are taught in school. a more intuitive solution: divide by b since b>0. a/b+b+a>/= 3 now we use the partial derivative test to find the minima. set both to 0, solve and you get the same answer.
casually you are watching tv, reading novel or playing suddenly you come to see this question a+b^2+b/a_>3b does this maths makes any sense our education system support to understand anything properly?
I know for a fact chess grandmaster MMA fighter and internet (CONTROVERSIAL 😂😂) personality Andrew Tate is not remotly aware of mathmatics but AI is master of us all this video kinda prpves it.
![Why you didn't learn tetration in school[Tetration]](http://i.ytimg.com/vi/Ea1_1aVfwl4/mqdefault.jpg)
In the last problem it is x^4 not x^3 so it becomes (x^4 + 1)/x >= 2x , sorry for the mistake
As x is strictly positive, we can multiply both sides by x, giving us that x^4 +1 >=2x^2, form this subtract 2x^2 form both sides leaving us with x^4-2x^2+1>= 0 this can be simplified into (x^2-1)^2 >=0 this is always true as x^2 is always greater then zero
Good job. You proved AM-GM for two members, you can actually apply it directly@@philippepare280
I was wondering because for 4/3 it's: 91/36>= 8/3=96/36 that doesn't sound correct
Do a linear algebra matrix question. Andrew tate talks about matrices.
😂
Underrated comment fr
He talks about "The Matrix"
This is happening
what rank would it be ?
Andrew Tate is a weirdly good math teacher
haha
@@PlanetNumeracy I love ChatGPT
we got Andrew Tate solved a Math Olympiad problem before gta6
if you thought his "Real world" discord was fire, wait until you see his "Real numbers" one
damn I actually felt like an alpha for a min😂😂😂
this isn't just playing with numbers, it's strategic warfare
For the first provlem we can transform it to
a + b^2 + b/a - 3b >= 0
We can multiply it by a since a > 0
a^2 + ab^2 + b - 3ab >= 0
Now I want to transform some part of left hand side into a quadratic form
a^2 + ab^2 + b^2 - b^2 + b - 2ab - ab >=0
a^2 - 2ab + b^2 + ab^2 - b^2 + b - ab >= 0
(a-b)^2 + b^2(a-1) + b(1-a) >= 0
Now I notice that (a-b)^2 is always non-negative, therefore we have to prove that the remaining part is non-negative
b(1-a) is the same as -b(a-1) which means that we have to prove that
b^2(a-1) - b(a-1) >= 0
b^2(a-1) >= b(a-1)
But here I have the issue because for 0
@user-ee4tz1zf4o your are close but not quite there. Even though what you have done is correct, the path you took does not lead you to the answer, you have to think of something else. A direct answer would be to apply AM-GM like mentioned in the video.
Here is my attempt at solving the last problem:
Multiply both sides by x so that we have:
x^4+1>=2x^2
Now move x^2 to the left side such that we get:
x^4-2x^2+1>=0
Split the middle term into two terms with coefficients that sum to -2 and multiply to +1:
x^4-x^2-x^2+1>=0
Factor by grouping:
x^2(x^2-1)-(x^2-1)>=0
Factor out (x^2-1):
(x^2-1)(x^2-1)>=0
Simplify:
(x^2-1)^2>=0
Since any real number squared is greater than or equal to zero, this inequality must be true for all x on the real domain.
Please let me know if I made a mistake, as it is very likely that I did; I don't have very much experience with proofs and would love to learn.
That's an original way of doing it. Well done, you can actully do it in a simpler way using AM-GM for two real positive numbers.
good one !
How did you get x^4 after multiplying by x?
@@avdomahmutovic4039 Look at the pinned comment on the video, which states that x^3 should actually be x^4
You did it wrong. When dealing with these kinds of inequalities you should check for when x is negative. You basically completely cross out that possibility. It should look something like this : (x^4+1)/x>=2x^2,(x^4-2x^2+1)/x>=0. From where you get all real positive numbers with the interval method. I apologize for my bad English
My idea of the solution.
a + b² + b/a ≥ 3b can be transformed to
ab² + (1-3a)b + a² ≥ 0
a>0 and b>0, so ab² > 0. Also, a²>0. If (1-3a)b would be ≥ 0 also, then the sum ab² + (1-3a)b + a² would be ≥ 0. But (1-3a)b≥0 only if a≤1/3. So I decided to determine under what values of _a_ this polynomial is ≥ 0.
Consider ab² + (1-3a)b + a² as a quadratic polynomial with respect to b. If the polynomial is ≥ 0, then it has 0 or 1 root (not 2), therefore its discriminant is ≤ 0.
D = (3a-1)²-4a³
D = -4a³ + 9a² - 6a + 1
So, if ab² + (1-3a)b + a² ≥ 0, then D≤0
-4a³ + 9a² - 6a + 1 ≤ 0
The usual strat for solving 3rd degree polynomial equations is to guess one of the root, which, in this case, is conveniently appeared to be a=1. Then use long polynomial division to factorize that.
-(a-1)²(4a-1) ≤ 0
a∈[0.25;+∞)
So. If a∈(0; 0.25), then ab², (1-3a)b and a² are positive, therefore their sum ab² + (1-3a)b + a² is positive.
If a∈[0.25;+∞), then ab² + (1-3a)b + a² has 0 or 1 roots, therefore ab² + (1-3a)b + a² is positive for all values of a,b or negative for all values of a,b. You can check that by randomly testing one of its values. For example, if a=2 and b=2, ab² + (1-3a)b + a² is 2. Therefore it's positive ∀b>0 and a≥0.25.
I've realized that it's too complex when 70% of this comment was done 😅. Finished it bc just didn't want that time to go to waste.
do you study math@@niko-niko-niiiiiiiii
I mean, cant complain, im literally learning Polynomials rn
For the final problem: Rearrange (since positive) x^4+1≥2x^2, now rearrange again, and we have x^4-2x^2+1≥0, but this is trivial since (x^2-1)^2≥0 for all x, equality holds when x^2=1, hence x=1
Well done!
Isn't the degree of the numerator equal to 3? It should be x^3+1 => 2x^2, right?
@@alitraore1496 check the pinned comment
@@alitraore1496 pinned comment
On the left side, multiply by a, add ab, and apply the AM-GM inequality. i think it should take around 7-50 secs to solve this.
this is a very creative and productive use of AI
(x^2-1)^2>=0
x^4-2x^2+1>=0
x^4+1>=2x^2 Since x is positive dividing both sides by x doesn’t change inequality.
(x^4+1)/x >=2x and equality happens when x=1
When I looked at this my first thought was a proof by induction. Start by noting a isn’t 0. Then if we take b=0 as the base case I can see conceptually the induction hypothesis of b=n+1 could be messy, but I think it would still work.
you can't apply induction on real numbers.
How did the matrices find me on youtube algorithm? Just when i thought i had escaped...
I don't think I fit the target audience of Andrew Tate but if calling things "battle tools" helps people get into math, I'm all for it!
Hello, thank you soo much for this materiel,
please can you offer more videos like that,we need it a lot , 🙏🙏
Keep going sir, my greetings
Ai is insane😂
@iro4201 shut up.🙄 . Go live on mars.
Just make it x^3+(1/x)>=2x and by Am-Gm,( x^3+(1/x))/2>=sqrt(x^2) which means that x^3+(1/x)>= 2x and equality holds when x^4=1 which means x =1.
x^3 + (1/x) is not equal to (x^3 + 1) so your solution is not true. Multiply both sides by x (x is not equal to 0) and you get x^3 -2x +1 >=0, you can get the roots of the equation by using rufinni with x=1 and the second grade equation formulla. You get that that equation is negative between the interval of 1 and 1.618, you can verify that if you put x=1.1 on the original question, it is not true
Since (x^2-1)^2 >=0
Because square of any real number gives a positive value.
So, x^4 - 2x^2 + 1 >= 0
x^4 + 1 >= 2x^2
Now since x>0 so,
(x^4 + 1)/x >= 2x
At x=1 LHS=RHS
Nice one ! if you enjoyed that one you must like this too : th-cam.com/video/fYKf5qFhw-E/w-d-xo.html&lc=Ugy3FxscSnwsNDweOKV4AaABAg
the andrew tate teens needed
AI Andrew tate
No this was actually him! are you dumb?
the solution feels a little random though it is probably what you are expected to use in a math test especially considering that AM-GM inequalities are taught in school.
a more intuitive solution:
divide by b since b>0.
a/b+b+a>/= 3
now we use the partial derivative test to find the minima.
set both to 0, solve and you get the same answer.
Except it's a/b + b + 1/a, champ.
a world where Andrew Tate is teaching math we would have flying cars
AI (Andrew Intelligence)
Hiw they do this AI with Andrew Tate voice?
No wonder this guy’s always talking about matrices-he does Olympiad math.
Nahh, how is this andrew tate more charismatic and wise than the real one?
Thanks Mr.Tate
How did you make this? Is there a text to speech website that has voices of famous people?
yes how??
AI
@@redgrengrumbholdt2671
Yeah but which AI
He better mention the Hoffman paccking puzzle.
casually you are watching tv, reading novel or playing suddenly you come to see this question a+b^2+b/a_>3b does this maths makes any sense our education system support to understand anything properly?
Where’s that next video? I find the inecuation shown at the end is wrong
check the pinned comment, it says : it is x^4 not x^3 so it becomes (x^4 + 1)/x >= 2x , sorry for the mistake
Wow!! I am pretty impressed
AI is getting scary
when andrew see this. "GEEKS BRUH" 🤣🤣🤣🤣🤣🤣🤣🤣🤣
the good ending andrew tate
Blud does NOT know that a and b should be positive real numbers to apply AM-GM🙏🙏😭
good point, but it was mentioned
Bro that's what literally what was in the problem
Bro that's what literally what was in the problem
? are u blind
Bro that's what literally what was in the problem
how did you make this?
Warriors of wisdom 😂😂😂😂
That was cool !
Why did it took me to solve it in 10 sec 😂...
Sorry I m an indian
How do you use Andrew tate's Voice?
can we not do this just by deduction?
what do you mean?
Ngl this would probably get kids interested.
The wisdom of artificial intelligence.
If g was a numerator, it would be..
top g !
Andrew Tate wouldnt know what an implicit variable was if it fell for his cam girl scam
00:01
This is incredible
*I love the internet ❤*
I know for a fact chess grandmaster MMA fighter and internet (CONTROVERSIAL 😂😂) personality Andrew Tate is not remotly aware of mathmatics but AI is master of us all this video kinda prpves it.
Check Mr beast working on a math Olympiad problem :
th-cam.com/video/ygl240Y3hjw/w-d-xo.html
which AI tool are you using ? can you create one like Robert Downey Jr. is solving this ?
perfect
sounds ai
Like this Andrew Tate better
Andrew Tate finally using the one braincel he has
"Arithmetic" as an adjective should be pronounced differently.
if this is begginer level then im fucked
No intonation though. But accent is decent Ig
not bad for a guy who doesn’t read books
Based
Isn't this false if a=-2000 and b=1?...Lol...I mean, or I'm just miscalculating...
check the pinned comment, it says : it is x^4 not x^3 so it becomes (x^4 + 1)/x >= 2x , sorry for the mistake
Shame on the audience that they give more views to a math video involving the name Andrew Tate rather than Ramanujan
First time on Internet? Jesus.
where are ramanujan ai dubbed videos?
isnt this ai
So funny, aren’t you data scientist ? 😮
Sounds like Ai
Nice ai
😂😂😂awesome
I laughed alot
That's why i hate math.
can we do diddy
no
das crazy
is this AI generated?
lol
Lol
Fake voice...
Made by AI...🙂
AI generated...Fake :)
Why Andrew Tate? It's cringe 😬