What is the minimum of this expression?
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- Опубліковано 16 жов 2024
- We look at a nice problem from the Southeast China Math Olympiad.
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8:11 yes, we can get this inequality other way, which is much more conventional.
We have a cubic polynomial p(x) = x³-ax²+bx-a, which has three real roots. Its graph therefore has a minimum and a maximum, or at least an inflection point, where both derivative is zero and value of polynomial is zero. The derivative p'(x) = 3x²-2ax+b and it must have two real roots (maybe, a double root), so its discriminant must be non negative. So we have: D=(2a)²-4*3*b = 4(a²-3b)≥0, which implies a²-3b≥0
I would never have thought of that, good intuition!
Nice
why it is necessary for p'(x) that it must has 2 roots?
@@bangstar719 Cubic p(x) has three roots when it has an inflection point and two extrema (one local maximum, one local minimum), and in one of which it has negative value, on the other a positive value. Those extrema lie exactly where p'(x)=0, so there must be two roots of a derivative for the cubic to have three roots.
Woah, that trick with dot products of parallel vectors is so nice!
Your thumbnail is 👍 👍!!!
Hi, can you make a video about proving an integral is nonelementary?
Can u make a similar DI setup for :
-Summation by parts (en.wikipedia.org/wiki/Summation_by_parts) ;
-Abel's summation formula (en.wikipedia.org/wiki/Abel%27s_summation_formula) ?
There's a mistake at 3:55, when you cube the inequality 3*cuberoot(a) less than a, you get 27*a less than a^3, not a^2.
Correct
But he solve it right away, clever guy
if you pay close attention this is actually a typo (and he misspoke) but the math is still correct
@@viper-sagi yes, that's true, but he didn't say yo us for his mistake, So we can learn wrong.
He had already finished the next step in his head by the time he started writing down the exponent.
for the dot product explanation, using Cauchy Swartz (which is literally the same thing) suffices. But the statement a^2+ b^2+c^2 》ab+bc+ac has many ways to prove.
Rearrangement inequality as well
Ohhhh! Thanks for pointing out that this is all Cauchy-Schwartz is saying. CS felt like this arcane fact, but the dot product argument was so much clearer.
If you want to be really thorough, you can note that the AM-GM inequality becomes an equality only when all the inputs are equal, which in this case gives r1 = r2 = r3 = sqrt(3)
Cant you solve this without AM and GM ?
Instead of dot product
We can write
(r1-r2)^2+(r2-r3)^2+(r3-r1)^2≥0
On expanding we get the required inequality
elegant!
In 8:30, knowing that b is positive from polynomial roots, we can find the minimal value much easier, using a^2/3>=b inequality:
(2a^3-3ab+3a)/(b+1)
>=(2a^3-3a*a^2/3+3a)/(a^2/3+1)
=3a>=9sqrt(3), which gives us the same potential minimum
8:55, what happened to coeff of a^3 that is changed from 2 to 3?
Just a typo.
@@emiltonklinga3035
3's and 2's are repeated typos here :)
That's a classic Michael Pen mistake, you can find those in many of his videos :)
Doesn't diminish his great content though.
I paused the video and have to answer the same, just a classic doctor Penn typo, I liked them they make me think and check
The a^2>3b inequality can be derived by imposing that the derivative of the third degree polynomial has two positive real roots, which is a necessary condition to the fact that the the polynomial has three roots
You also have to check that the final value of a and b make the poly one with 3 positive roots
Exactly, that is a hole at the end of the video. The polynom actually factors as (x-sqrt(3))^3.
Yeah, it needs clarification, we either allow degenerate roots or just say poly has only positive real roots. Otherwise we should use strict inequality, minimum will act as a limit, but its value remains the same, it just can't be reached.
Good motivation for me, as i am starting off with AP-GP in a few weeks. Kudos to this beautiful channel
Always vieta when dealing with roots!
(r1-r2)^2>=0 so 1/2*r1^2+1/2*r2^2>=r1*r2, same for r1,r3 and r2,r3. Combine all three.
compi does it too:
*Minimize[{(2 a^3 - 3 a b + 3 a)/(b + 1), x^3 - a x^2 + b x - a == 0, x > 0}, {a, b}, Reals] // FullSimplify*
3:49 I have a quick question. Genuine question. Why cubing both side will result in RHS becoming a^2 instead of a^3?
It does not. He just made a mistake.
@@alexandermendez6083 Thanks Alexander.
@@adminguy but he corrected it in the very next step 👍
I should stop solving based on the thumbnail. I figured there was no minimum until I clicked and saw the restriction
The second inequality to bound b can be found easier using basic inequalities. Just use the arithmetic - quadratic inequality on the sum of squares of roots:
sqrt((r1²+r2²+r3²)/3) >= (r1+r2+r3)/3
by squaring both sides, replacin sum of squares of roots with a²-2b and replacin sum of roots with a you get that b
Mistake at 8:40 (it should be 2a^3)
4:34 Michael Sussy...
mugus
A important note on his usage of the dot product inequality is that it is only true if r1 r2 and r3 are not equal, since otherwise it would still be the same vector and thus equal, but we know the roots are not equal because of they were the cubic would be a perfect cube, but given that it says the roots are all positive, there is no way -a is a perfect cube, since 3 positive no.s multiplied gives positive.
At 9:40 the factorisation is wrong. Should be 3a(a^2+2)
For the dot product one you can try the rearrangement inequality. It’s really useful and can be used to prove things like AM-GM and Cauchy-Schwarts.
Love your channel, but that red chalk is difficult to read on my phone screen 2:50
How at 9:00 2a^3 become 3a^3 in the numerator?
EDIT: i figured out is only a type error
Just a typo.
He compensates at 9:27 with another error which gets him back on track.
@@kevinmartin7760 legend
The dot product inequality is basically the rearrangement inequality
I think you made a mistake at 3:58, you Shall not square both sides, you should cube them
He actually did. The 27a ≤ a² was just a typo and should of course have been 27a ≤ a³. You see that in the next step when he divides by a and achieves 27 ≤ a².
The result a >= 3 sqrt(3) is still correct, since the inequality a^2 >= 27 follows from a^3 >= 27 a, since a must be greater than 0 (since it is the product of positive roots).
Love your videos, but can you check the factoring at 9:33? I can factor out 3a, but not 2a.
He typoed the 3a³, it's actually 2a³.
For a finite list of positive numbers, you can prove the inequality by ordering the numbers then using a greedy algorithm. I.e., to maximize the sum of 1 number, we want the biggest number multiplied by the largest factor (itself) then repeat on the remaining numbers via induction to get the sum of squares. I recall this proof being used in competition prep to show that the maximal sum of products of two positive ordered sequences is to pair the terms using the same numbering, while to minimize the sum, you simply reverse one of the sequences
Solving over the integers, 6 is perfect = 1 + 2 + 3 = 1*2*3, and the min in this case is 21.
Hi,
For fun:
0:13 : "our goal is to",
4:18 : "ok, great",
8:22 : "our goal is to".
12:09
I will have to time travel much faster :(
How is this comment 5 days old🤨🤨
Voodoo magic?
@@timewaster0123 Time travel, yep
The minimum 9sqrt3 is for one positive root with multiplicity 3 and not for 3 positive roots
The way I read it, the problem did not state that the roots had to be distinct, just real and positive. Although one might be of the view that multiple (multiplicitous?) roots are a single root, perhaps the wording should have been that "all the roots are positive real) rather than "has 3 positive real roots".
If you require all roots to be distinct I suspect there is no minimum for the expression, just an infimum (whose value is 9sqrt(3)).
@@kevinmartin7760 yes but when you say 3 it may confuse at least i was confused so it is more clear if you say all the roots are positive or to say including roots with multiplicity
Yes. The problem does not say that the roots must be distinct. I believe that if we force the roots to be distinct, there is no minimum value.
@@kevinmartin7760 and yes there is no minimum if they are distinct so i break my head to figure it out until i saw the solution
@@kevinmartin7760 maybe can have a solution to three diferent positive roots, but i thought would need to use complete cubic root formula, but i'm not sure
I went to the discriminant of the cubic which took me way longer to work out and then ended up being a much more complicated inequality that I couldn't use...
HOMEWORK : Compute the length of the interval of values x for which 1/(x + √x) + 1/(x - √x) ≥ 1.
SOURCE : ARML Local 2014
Wait, was this video ready to watch 4 days ago?
Bigger than or equal to what?
@@falquicao8331 Bigger than or equal to 1.
SOLUTION
*2*
1/(x + √x) + 1/(x - √x) = 2x/(x(x-1)). The domain of the function is all positive x except 1. For x between 0 and 1, the denominator is negative, so the function is negative. For x greater than 1, the function is strictly decreasing. It equals 1 when 2x/(x(x-1)) = 1, or x = 3, so the interval is (1,3) which has length 2.
Where are you from? ...another galaxy....what time is in your galaxy now?
When I saw Dat a=r1+r2+r3=r1r2r3, I remembered Dat when a+b+c=abc, one or the only solution is √3
assuming a=b=c.
So a would be 3√3 and b √3√3+√3√3+√3√3=9
Dat was wat u got in the inequality for a and if u had replace the a in the b inequality, and idk any of that complex stuff
Hello
... and what if such values for a and b (11:30) didn't exist?
Could you please solve an example for that condition?
If you use the fact that (r1-r2)^2+(r2-r3)^2+(r3-r1)^2 is bigger than or equal to 0 you can prove a^2 is bigger than or equal to 3b
I suggest using AM GM inequality again when trying to substitute the sum of squared value
@8:48, where did you get the 3 coefficient in 3a^3? it was a 2 before.
Just a typo.
Use of dot product was good.
Whats the biggest it can be while being smaller than the next biggest it could be
For the inequality that you proved using dot product , here's the simpler way
From AM>=GM
r1^2 + r2^2 >= 2r1r2
r2^2 + r3^2 >= 2r2r3
r1^2 + r3^2 >= 2r1r3
Add all these and you get the result
But with a=3sqrt(3) and b=9 the polinomial expresión don't have 3 positive real roots, it just has x=sqrt(3)
In algebra the poly'al of nth degree always has n complex roots. Some (even all) of them could coincide, they sometimes referred to as a root of multiplicity greater than one. For example, x⁵=0 has 5 roots: 0, 0, 0, 0, 0, or, a root 0 of multiplicity 5.
At 10:10 factor 3a not 2a. Thanks
Why not just use Legrange multipliers?
Because the condition on the cubic polynomial is not an equality.
lol, was watching on mute and spent about 10 minutes how you got (r1+r2+r3)/3 = 9/3 before I was able to take off mute and realized you wrote a/3
"And I think there's maybe an easy way to do this"
Isn't that just the rearrangement inequality?
From when he stopped saying that's a gud place to start and started saying that its a gud place to stop ‽
I have no idea !
Sir you make a mistake at 3:58 betwen a square to a cube. U r humain like us .never mind.thanks for you.we miss your live stream
9:30 so 3/2=2?
No
3/2=1.5
A math professor was teaching calculus in college. At one point he was making mistakes in writing on the blackboard. One student pointed out the mistake to which the professor replied, "don't focus on what I am writing, focus on what I am saying".
This continued and after some time he starts making mistakes in what he was saying too. The perplexed student pointed it out again. The professor said, "Focus not on what I am writing, bother not about what I am saying but rather feel what I am feeling".
So, basically with this video I've learned that 2 = 3
Why do I find some math problems like this one just irritating?
The values for a and b given at the end yield a cubic equation in x with only 1 positive real root of multiplicity 3.
I would interpret this as being invalid given how the question is worded. Then again, I think if we don't allow it then there is no minimum and just an infimum.
After you get a≥3√3 and a²≥3b i Just put 3√3 in the fraction because a Is only in the numerator so the small a Is the small the fraction Will be. Then After some Easy calculation i get 9√3(19-b)/(b+1) wich Is Always decreasing, so i Only find when a²=3b wich is b=9, and put this back in the fraction we get 9√3
The denominator needs to have grouping symbols around it: (b + 1)
@@robertveith6383 thank you!
How can you assure that one cannot find some other inequalities to use so that the minimum values that he finds gets smaller than what you've got.
I can't believe you didn't mention Vieta's fomulas when expanding (x-ri).