I solved this inequality by the time I was interesting in math competitions by substituting a, b and c by tan(alpha), tan(beta) and tan(gama). That's a very nice trick to work with. With angles in [0,pi/2] tangent function assumes any positive real value and we can take advantage of the denominators
Hello pls how can i know which trig subsistution to do ? Tan or sin or cos or what , and when ? and how can i know in which interval i place the angles ?
@@shreddedtwopack6625 I would assume plugging in tan(beta) into the denominator of the first term yields sec^2(beta) but I'm not sure how it would follow from there
@Alem Memic I was thinking along this line too, because usually the method he is talking about comes up with the condition that a+b+c = abc. But then this means that a,b and c all have to be non-zero, because if any of them are while the other two are not (ie nonzero), then you would have (for example say a = 0) 0 + b + c = (0)*b*c -> b+c = 0 which is impossible because b and c are real positive numbers. Therefore the condition fails and you would need the extra assumption that a,b,c > 0 for this video's problem. Plus this method is used to show that something is less than 3/2, eg the 1998 Korean Olympiad problem Show for positive a,b,c such that a+ b + c = abc that 1/sqrt(1 + a^2) + 1/sqrt(1+b^2) + 1/sqrt(1+c^2)
a+b+c=3 => smallest of the a,b,c is =1 hence a,b,c=tanx, tany, tanz substitution is wrong in interval (0,pi/2). also that's a different question where each term was function of 1 variable only.
I click you videos due to the great content you have consistently posted for years. Don't worry about titles and thumbnails; your content and earned reputation are strong enough to propel you forward.
I agree. a >= 0, b >= 0, c >= 0 is not sufficient since eg. a=b=c=1/2 gives 3/4 for the sum of squares, which is clearly less than 3/2. Using a + b + c = 3, we can replace a^2 + b^2 + c^2 with a^2 + b^2 + (3 - a - b)^2.
Strictly speaking, this inequality is called "cyclic", not "symmetric". An inequality of 3 variables F(a,b,c) is called symmetric iff all F(a,b,c), F(a,c,b), F(b,a,c), etc. (any order of the triplet a,b,c) hold true. On the other hand, F(a,b,c) is called cyclic iff all F(a,b,c), F(c,a,b), F(b,c,a) (note the cyclical "transition") hold true.
@@wannabeactuary01 No, not at all. If you swap b and c, you get: F(a,c,b) = a/(b^2+1) + c/(b^2+1) + b/(a^2+1), which evaluates to 1.2 with (a,b,c)=(1,2,0). The inequality breaks down immediately.
@@MrPerfectstrong F(a,c,b)=a/(c^2+1)+c/(b^2+1)+b/(a^2+1). You are right that this is not the same term, but the inequality is still true, obviously. You just changed some variables, after all.
Correct. And somehow i find the expression, "ab + bc + ac" annoying. Come on people, "ab + bc + ca", has a nice ring to it, cyclic, isn't it? So... ca, NOT ac!!
Maximizing ab+bc+ca over constraint a+b+c=3 is easily done using Lagrange multipliers, which are extremely powerful and would love to see you use more in the future
This is an overkill for this problem? set a = 1+ x, b = 1+y, c = 1+z. Then a^2 + b^2 + c^2 = 3 + x^2 + y^2 + z^2 (2x + 2y +2z = 0 since x+y+z = 0) >= 3 = a + b + c.
0:26 Well the style is good but maybe this thumbail just missed the fact we dont want values for a,b,c but a proof of the inequality. Maybe add « if » and « then » would help? 9:28 Good Place To Stop
I always enjoy the titles along the lines of « a tricky inequality with a beatiful proof » as I’m personaly always looking for that type of content regarding math on youtube (i.e. The beautiful proofs)
1. I think your thumbnail and title were great here. The title is short and descriptive, but also provokes curiosity (or at least, it did for me...) 2. Several comments have pointed out that the inequality used near the end is not quite right, but that it follows from the RMS-AM inequality. I wanted to add that this also follows from the Cauchy-Schwarz Inequality, applied to the vectors (1,1,1) and (a,b,c)
@@Nrbebbebh3 Search in google, of course! And if you want to find problems for it, the only source I know is in Vietnamese, so maybe I can translate some of the problems for you.
en.wikipedia.org/wiki/Generalized_mean you should use it on the last board: sqrt((a*a + b * b + c * c) / 3) >= (a + b + c) / 3 = 1 squaring both sides gives us desirable result: (a * a + b * b + c * c) >= 3 in common case, only correct next inequality: (a * a + b * b + c * c) >= ((a + b + c)** 2) / 3
In Vietnamese maths we typically call this technique "reverse inequality sign AM-GM". Basically looking at the equation we cannot just use AM-GM in the denominator, say a^2+1>=2a, because then the inequality sign is reversed making the problem seemingly impossible from this point. Instead, we use the technique, as shown in this video of Professor Micheal Penn.
Another elegant solution would be : Express each term of inequality as , (a/b)(b+1/b) + (b/c)(c+1/c) + (c/a)(a+1/a ) and apply AM >= GM , you will directly get the result.
To perhaps include a graphic on the thumbnail for inequality-type problems, if the problem involves 3 variables, as in this case, you could display a plot of the function subject to the restriction and include a point which represents the value with which it will always be bounded by.
Title suggestion: "Unbelievable! One hard inequality quickly defeated by two easy inequalities". You are a top-notch teacher, Michael! Take it easy on these climbing workouts, I guess. Keep up the great work.
I think this solution needs to start with the obvious values for which you get equality. This can be used to justify some of the choices for the used inequalities. Either you can say, this is cyclic so we can check if a=b=c gives exact equality - it does and moreover a=b=c=1 since their sum is 3. Therefore (y-1)^2>=0 is a good starting point. Moreover since a=b=c is exact solution this strongly suggests arithmetic mean/geometric mean/quadratic mean inequalities
1. I think that putting the inequality on a thumbnail is enough 2. 8:35 There's no name for this inequality beacuse its wrong. The actual inequality is a^2 + b^2 +c^2 >= 3((a+b+c)/3)^2, it called Inequality of root mean square and arithmetic means, or simpler RMS-AM inequality. You were just lucky that a+b+c=3, and thats why it works.
Came to the comments to find out what happened to the left hand, shocked to see that no-one has asked that question! So... What is with the bandage on the left hand?
As many people have said, it can be implied from Cauchy's inequality, which states for real numbers a1, a2, ..., an and b1, b2,...,bn , a1*b1 + a2*b2 +... + an*bn
@@vardhanr8177 Pleasure. Yeah, Cauchy's inequality, the Arithmetic mean-Geometric Mean (AMGM) and Jensen's inequality are all extremely important and good ones to know (as well as Cauchy-Schwarz). If you want an excellent book for doing these, I recommend "The Cauchy-Schwarz Masterclass" by J. Michael Steele. The problems are long and super difficult but it really grounds how to do/manipulate a lot of these inequalities which is really nice and he really goes through the intuition of where these inequalities came from and how they were found. Also, because he's a probabilist by training, there are some amazing problems that you would rarely see in a standard probability course beyond stuff like Chebyshev's bound that he pulls from papers or inequalities used in the actual every-day working/research field of mathematical statistics or probability.
better aprroach use titu lemma and solve it under 3 lines ofcourse there will me some manupulation but it will be solved easily i cant type all step but try that approach too
Have you ever considered doing some of the ridiculous diophantine equations that look really simple? Solve a/(b+c) + b/(a+c) + c/(a+b) = 4, a,b,c all positive integers is a great example. Looks simple enough at a casual glance. Yet this problem is an absolute beast, there are probably fewer than a quarter million people alive who can solve it. If you want to be really fiendish, post it with cute fruits replacing the symbols, like all of those simple problems that do the rounds on Facebook. There's your clickbait thumbnail for it
@@MaserXYZ Yeah, but you will require the following machinery to solve it: - Brute force finding a solution that fails the 'all positive' criteria (not trivial but no absolute values exceed 12) - Transforming diophantine equations into problems involving the group of rational points of an elliptic curve via Weirstrauss transformation - Repeating the elliptic curve group operation until you transform the near-solution into an actual solution The smallest answer has about 80 digits for each of a, b and c. I wasn't exaggerating when I said not many can solve it. My honours thesis was about number theory and elliptic curves - this wasn't enough for me to solve the problem, but it did let me (mostly) follow a solution. The only person at the Uni of Sydney who could have solved this when I was a student there was my thesis supervisor.
Your upper bound estimate of a quarter million people being able to solve this sounds absurdly too large, given that a smallest value has about 80 digits.
@@robertveith6383 you'd never find it by brute force but people with a master's level degree narrowly in number theory will all get this. Some honours level people as well.
I probably amn’t ‘qualified’ to answer your question fully since I watch all your videos regardless. If you uploaded 10 minutes of a black screen and titled it tomatoes I’d watch.
Using it directly on the given quickly turns into algebra nightmare, but using it to maximizing ab+bc+ca constrained to a+b+c=3 would quickly result in single critical point a=b=c=1. The rest is just a matter of checking the boundary a=0 which can be done using much simpler techniques (am-gm or single variable calculus)
Can't this be proved by inspection in about 30 seconds just by looking at all possible values: (0,0,3), (0,1,2), and (1,1,1) (due to symmetry)? Not as much fun, but direct...
It's not super clear, but nowhere is it mentioned that a,b,c are integers. They're just bigger than or equal to 0. If it was just integers, you could brute force it, but more tricks are needed if they're not integers.
So satisfying. Thank you. But I think the last part was obvious and it was unnecessary to do. Am I right?! ( The first page, or the title of video is wrong! We have smaller than 3/2 instead of bigger than 3/2)
I personally like titles which describe the content of the video succinctly. "A tricky inequality" is probably a little broad. As for thumbnails, I think your current style is actually really nice. It shows the problem to be solved, which I think is the biggest bait of a video, whilst also not being a lie or dishonest.
for titles, always start the title with the main word - INEQUALITY - followed by the branch about the content. That way all your videos can be easily searched and listed?
F*k it, looks like Mike has come up with a good strategy of making the video clickable. Just show the wrong inequality sign in the thumbnail. Well, I felt for it once, but next time I will report the video for misinformation.
Title suggestions : "You'll never guess the answer to this equation!!!", "I tried to solve this problem, then the unexpected happened!!!", "Only 1% of people will be able to understand this!", "The 5th step of this proof is just AWESOME!"... :D
open answer: "a symmetric inequality THE EGYPTIANS LEFT US or FROM THE DAWN OF TIME or KING EDWARD THE THIRD WAS VERY FOND OF or FOUND IN THE CORNER OF THE HOLY SHROUD OF TURIN or ALLOWING DOGS AND CATS TO LIVE TOGETHER WITHOUT THE MASS HYSTERIA (CIT.)"
I solved this inequality by the time I was interesting in math competitions by substituting a, b and c by tan(alpha), tan(beta) and tan(gama). That's a very nice trick to work with. With angles in [0,pi/2] tangent function assumes any positive real value and we can take advantage of the denominators
Hello pls how can i know which trig subsistution to do ? Tan or sin or cos or what , and when ?
and how can i know in which interval i place the angles ?
nice approach
@@shreddedtwopack6625 I would assume plugging in tan(beta) into the denominator of the first term yields sec^2(beta) but I'm not sure how it would follow from there
@Alem Memic I was thinking along this line too, because usually the method he is talking about comes up with the condition that a+b+c = abc. But then this means that a,b and c all have to be non-zero, because if any of them are while the other two are not (ie nonzero), then you would have (for example say a = 0) 0 + b + c = (0)*b*c -> b+c = 0 which is impossible because b and c are real positive numbers. Therefore the condition fails and you would need the extra assumption that a,b,c > 0 for this video's problem. Plus this method is used to show that something is less than 3/2, eg the 1998 Korean Olympiad problem
Show for positive a,b,c such that a+ b + c = abc that
1/sqrt(1 + a^2) + 1/sqrt(1+b^2) + 1/sqrt(1+c^2)
a+b+c=3 => smallest of the a,b,c is =1 hence a,b,c=tanx, tany, tanz substitution is wrong in interval (0,pi/2). also that's a different question where each term was function of 1 variable only.
I click you videos due to the great content you have consistently posted for years. Don't worry about titles and thumbnails; your content and earned reputation are strong enough to propel you forward.
8:45 I don't think a^2+b^2+c^2 is always bigger than or equal to a+b+c, but that's true when a+b+c=3
so much for "well known" inequality!
I agree. a >= 0, b >= 0, c >= 0 is not sufficient since eg. a=b=c=1/2 gives 3/4 for the sum of squares, which is clearly less than 3/2. Using a + b + c = 3, we can replace a^2 + b^2 + c^2 with a^2 + b^2 + (3 - a - b)^2.
@@emanuellandeholm5657 how would that help?
@@amirb715 Not sure, but it should help with some clever manipulation.
@@amirb715 I tried lookout up. There’s no such general inequality
Strictly speaking, this inequality is called "cyclic", not "symmetric". An inequality of 3 variables F(a,b,c) is called symmetric iff all F(a,b,c), F(a,c,b), F(b,a,c), etc. (any order of the triplet a,b,c) hold true. On the other hand, F(a,b,c) is called cyclic iff all F(a,b,c), F(c,a,b), F(b,c,a) (note the cyclical "transition") hold true.
I think in this case symmetric :-)
@@wannabeactuary01 No, not at all. If you swap b and c, you get: F(a,c,b) = a/(b^2+1) + c/(b^2+1) + b/(a^2+1), which evaluates to 1.2 with (a,b,c)=(1,2,0). The inequality breaks down immediately.
@@MrPerfectstrong F(a,c,b)=a/(c^2+1)+c/(b^2+1)+b/(a^2+1). You are right that this is not the same term, but the inequality is still true, obviously. You just changed some variables, after all.
Correct. And somehow i find the expression, "ab + bc + ac" annoying. Come on people, "ab + bc + ca", has a nice ring to it, cyclic, isn't it? So... ca, NOT ac!!
Maximizing ab+bc+ca over constraint a+b+c=3 is easily done using Lagrange multipliers, which are extremely powerful and would love to see you use more in the future
This is an overkill for this problem? set a = 1+ x, b = 1+y, c = 1+z. Then a^2 + b^2 + c^2 = 3 + x^2 + y^2 + z^2 (2x + 2y +2z = 0 since x+y+z = 0) >= 3 = a + b + c.
For the end you can use (a+b+c)^2>=3 (ab+bc+ca)
0:26 Well the style is good but maybe this thumbail just missed the fact we dont want values for a,b,c but a proof of the inequality. Maybe add « if » and « then » would help?
9:28 Good Place To Stop
Wtf , you're 24/7 on youtube?
@@thatmathnerdteachesphysics5180 😂
@@thatmathnerdteachesphysics5180 notifications, my friend, notifications
@@utkarshsharma9563 but even to see a notification you must be using your phone right?
@@thatmathnerdteachesphysics5180 yeah, I mean maybe he keeps the phone nearby all the time
This is a very powerful techniques
The main trouble we face is
b^2+1>=2b
and this implies
1/(b^2+1)
cauchy schwarz inequality can significantly make the question easier but I liked ur method too.
I always enjoy the titles along the lines of « a tricky inequality with a beatiful proof » as I’m personaly always looking for that type of content regarding math on youtube (i.e. The beautiful proofs)
1. I think your thumbnail and title were great here. The title is short and descriptive, but also provokes curiosity (or at least, it did for me...)
2. Several comments have pointed out that the inequality used near the end is not quite right, but that it follows from the RMS-AM inequality. I wanted to add that this also follows from the Cauchy-Schwarz Inequality, applied to the vectors (1,1,1) and (a,b,c)
That will give you :
(a^2 + b^2 +c^2 )(1+1+1)>= (a + b + c ) ^2
(a^2 + b^2 +c^2 )>= ((a + b + c ) ^2 )/3
Wich not equivalent to what he said
If you want to practice doing inequalities like this, I recommend searching for AM-GM inequalities and the CBS inequalities.
Search in where? Where can we find problems with solution? (Except aops)
@@Nrbebbebh3 Search in google, of course! And if you want to find problems for it, the only source I know is in Vietnamese, so maybe I can translate some of the problems for you.
In the thumbnail this video currently has the inequality sign is facing the wrong way
I think your thumbnails work great. You deliver on your promises, and I find that great!
My math teacher just gave me this problem and I completed it right when this video just came out! What a coincidence :D
You did it right? (You proved it?)
For which class?
it's simplier to show that (a+b+c)^2=9 >= 3(ab+bc+ac) using a^2+b^2+c^2>=ab+bc+ac
=> ab+bc+ac -1/2(ab+bc+ac) >= - 3/2
=> 3-1/2(ab+bc+ac) >=3/2
en.wikipedia.org/wiki/Generalized_mean you should use it on the last board:
sqrt((a*a + b * b + c * c) / 3) >= (a + b + c) / 3 = 1
squaring both sides gives us desirable result: (a * a + b * b + c * c) >= 3
in common case, only correct next inequality: (a * a + b * b + c * c) >= ((a + b + c)** 2) / 3
thank you
In Vietnamese maths we typically call this technique "reverse inequality sign AM-GM".
Basically looking at the equation we cannot just use AM-GM in the denominator, say a^2+1>=2a, because then the inequality sign is reversed making the problem seemingly impossible from this point.
Instead, we use the technique, as shown in this video of Professor Micheal Penn.
Another elegant solution would be :
Express each term of inequality as , (a/b)(b+1/b) + (b/c)(c+1/c) + (c/a)(a+1/a ) and apply AM >= GM , you will directly get the result.
To perhaps include a graphic on the thumbnail for inequality-type problems, if the problem involves 3 variables, as in this case, you could display a plot of the function subject to the restriction and include a point which represents the value with which it will always be bounded by.
Title suggestion: "Unbelievable! One hard inequality quickly defeated by two easy inequalities". You are a top-notch teacher, Michael! Take it easy on these climbing workouts, I guess. Keep up the great work.
😂😂
And that's a good place to stop, I like it when he says that
Evidently the way to thumbnail these is to get the thumbnail clearly wrong (it said = 3/2)
Honestly the reason why I clicked on the video 😅
Just after another video suggested 2 days before seen Michael's one, I realized this
I think this solution needs to start with the obvious values for which you get equality. This can be used to justify some of the choices for the used inequalities. Either you can say, this is cyclic so we can check if a=b=c gives exact equality - it does and moreover a=b=c=1 since their sum is 3. Therefore (y-1)^2>=0 is a good starting point. Moreover since a=b=c is exact solution this strongly suggests arithmetic mean/geometric mean/quadratic mean inequalities
I seem to find alternative, simpler solution: noting that b + 1/b >= 2b / (b^2 + 1), we represent a / (b^2 + 1) as a / (b * (b + 1/b)), and find that a / (b^2 + 1) >= a(b^2+1) / 2b^2 = 1/2( a + a/b^2)
a/(b^2 + 1) + b/(c^2 + 1) + c/(a^2 + 1) >= 1/2 ( a + b + c + c/a^2 + a/b^2 + a/c^2) >= 1/2 ( 3 + c/a^2 + a/b^2 + a/c^2) >= 3/2
Why is it always that when the function and constraint are symmetric, the function is optimized when all variables are equal (a=b=c)?
1. I think that putting the inequality on a thumbnail is enough
2. 8:35 There's no name for this inequality beacuse its wrong. The actual inequality is a^2 + b^2 +c^2 >= 3((a+b+c)/3)^2, it called Inequality of root mean square and arithmetic means, or simpler RMS-AM inequality. You were just lucky that a+b+c=3, and thats why it works.
I think it's RMS and AM
the AM-GM inequality
@@2070user no its not
@@wojteksocha2002 wait I think it is the somewhat generalized AM-GM mean not the normal one
Agreed. Its not always true. If a, b & c are less than 1 then Michael's inequality is false. It happens to work here as a+b+c=3.
I think it would be mor clickable if you didn't reveal answer to begin with and pose question is it greater or less?
Just to be picky, aren't you allowing a division by 0 around 5:00?
If we go this way we should distinguish the cases where a=0, a=b=0, etc...
Yeah, he is without knowing it, cuz I think that he forgot that a, b, and/or c can = 0.
even if y=0 the inequality still holds
Came to the comments to find out what happened to the left hand, shocked to see that no-one has asked that question! So... What is with the bandage on the left hand?
When we were trying to substitute y.y+1 we used (y-1)squared, would it be wrong to use (y+1)squared?
Can someone please explain to me how a ^ 2 + b ^ 2 + c ^ 2 >= a + b + c (referring to 8:46)?
As many people have said, it can be implied from Cauchy's inequality, which states for real numbers a1, a2, ..., an and b1, b2,...,bn ,
a1*b1 + a2*b2 +... + an*bn
@@ummwho8279 ohh okay... thank youu... (but i'm going to have to read up on Cauchy's inequality... because i don't know that)
@@vardhanr8177 Pleasure.
Yeah, Cauchy's inequality, the Arithmetic mean-Geometric Mean (AMGM) and Jensen's inequality are all extremely important and good ones to know (as well as Cauchy-Schwarz). If you want an excellent book for doing these, I recommend "The Cauchy-Schwarz Masterclass" by J. Michael Steele. The problems are long and super difficult but it really grounds how to do/manipulate a lot of these inequalities which is really nice and he really goes through the intuition of where these inequalities came from and how they were found. Also, because he's a probabilist by training, there are some amazing problems that you would rarely see in a standard probability course beyond stuff like Chebyshev's bound that he pulls from papers or inequalities used in the actual every-day working/research field of mathematical statistics or probability.
@@ummwho8279 okay, thank youu
better aprroach use titu lemma and solve it under 3 lines ofcourse there will me some manupulation but it will be solved easily i cant type all step but try that approach too
This inequality is not symmetric, as switching a and b for example changes it. It is called a "cyclic" inequality.
Have you ever considered doing some of the ridiculous diophantine equations that look really simple?
Solve a/(b+c) + b/(a+c) + c/(a+b) = 4, a,b,c all positive integers is a great example. Looks simple enough at a casual glance. Yet this problem is an absolute beast, there are probably fewer than a quarter million people alive who can solve it.
If you want to be really fiendish, post it with cute fruits replacing the symbols, like all of those simple problems that do the rounds on Facebook. There's your clickbait thumbnail for it
Does this one even have a solution? I tried to solve it but ran into dead ends (no sols) *everywhere*.
@@MaserXYZ ruclips.net/video/Ct3lCfgJV_A/видео.html
@@MaserXYZ Yeah, but you will require the following machinery to solve it:
- Brute force finding a solution that fails the 'all positive' criteria (not trivial but no absolute values exceed 12)
- Transforming diophantine equations into problems involving the group of rational points of an elliptic curve via Weirstrauss transformation
- Repeating the elliptic curve group operation until you transform the near-solution into an actual solution
The smallest answer has about 80 digits for each of a, b and c.
I wasn't exaggerating when I said not many can solve it. My honours thesis was about number theory and elliptic curves - this wasn't enough for me to solve the problem, but it did let me (mostly) follow a solution. The only person at the Uni of Sydney who could have solved this when I was a student there was my thesis supervisor.
Your upper bound estimate of a quarter million people being able to solve this sounds absurdly too large, given that a smallest value has about 80 digits.
@@robertveith6383 you'd never find it by brute force but people with a master's level degree narrowly in number theory will all get this. Some honours level people as well.
For example, if the parameter "d" is involved in the inequality the title could be "how Big is this d"
Sus
I probably amn’t ‘qualified’ to answer your question fully since I watch all your videos regardless. If you uploaded 10 minutes of a black screen and titled it tomatoes I’d watch.
lol
In the initial thumbnail of the video there is = 3/2
Can someone let me know which competition this is from? Maybe I missed if it’s mentioned in the video..
I call them "A-B-C vicious ties"
"js x better than y?" feels like a good idea
How about an inequality/climbing mashup entitled “outward bound… from below”?
(Or something along those lines)
Is this possible using the method of Lagrange multipliers? You have an objective function and you are trying to maximize it given a constraint.
Using it directly on the given quickly turns into algebra nightmare, but using it to maximizing ab+bc+ca constrained to a+b+c=3 would quickly result in single critical point a=b=c=1. The rest is just a matter of checking the boundary a=0 which can be done using much simpler techniques (am-gm or single variable calculus)
The inequality u r looking for is RSM AM inequality
The well-known inequality is just quadratic mean >= arithmetic mean, squaring both sides.
a^2 + b^2 + c^2 >= a+b+c
isnt always true tho
@@شبلالإسلام-ظ6ض But the inequality is a^2 + b^2 + c^2 >= (a+b+c)^2 / 3, not the one that you mentioned.
thanks teacher
4:45 it's worth a while to check for y = 0. Otherwise, this trick doesn't work.
It doesn't matter, I see your name I click.
Even if the title would be "anyone who clicks this is something bad"
lol
What about Muirhead? Anyone wanna give a try?
Nice content
Cauchy-Schwartz inequality, followed by rearrangement inequality.
Lagrange
Can't this be proved by inspection in about 30 seconds just by looking at all possible values: (0,0,3), (0,1,2), and (1,1,1) (due to symmetry)? Not as much fun, but direct...
It's not super clear, but nowhere is it mentioned that a,b,c are integers. They're just bigger than or equal to 0. If it was just integers, you could brute force it, but more tricks are needed if they're not integers.
Hi l’inégalité est vraie pour tout a, b et c si et seulement si a+b+c>=(5/3)^2, ce qui est le cas ici
Sorry if this isn't helpful, but the best title and thumbnails are the ones that are not click _bait._
Title: Symmetric Shortcuts - Inequality 1
Don't worry about the clickability Michael, I will always click your videos!
the thumbnail is wrong, the inequality sign should be reversed.
Unequal symmetry > symmetric inequality
Thumbnail: move all variables at LHS/RHS at say " {expression} is Positive/Negative". In such a way no inequality signs needed
So satisfying.
Thank you.
But I think the last part was obvious and it was unnecessary to do. Am I right?!
( The first page, or the title of video is wrong! We have smaller than 3/2 instead of bigger than 3/2)
It's so amazing form Cambodia 🇰🇭
Title: Symetrical fixed questions
I love inequality probkems!!!!
I personally like titles which describe the content of the video succinctly. "A tricky inequality" is probably a little broad. As for thumbnails, I think your current style is actually really nice. It shows the problem to be solved, which I think is the biggest bait of a video, whilst also not being a lie or dishonest.
how about this title : « this is a tricky inequality with a beatiful proof » ; avoid it !!
I mean the way to get clicks is to do clickbait. “Math professors tackles this IMPOSSIBLE problem” or whatever
for titles, always start the title with the main word - INEQUALITY - followed by the branch about the content. That way all your videos can be easily searched and listed?
your problem is different on the thumbnail lol ( it says less than or equal to)
F*k it, looks like Mike has come up with a good strategy of making the video clickable. Just show the wrong inequality sign in the thumbnail. Well, I felt for it once, but next time I will report the video for misinformation.
Title suggestions : "You'll never guess the answer to this equation!!!", "I tried to solve this problem, then the unexpected happened!!!", "Only 1% of people will be able to understand this!", "The 5th step of this proof is just AWESOME!"... :D
Inequalities gone wrong
Modular arithmetic can solve (almost) all your problems!
1 simple trick to solve inequalities!!!!!
So like Presh Talwalker then.
@@anonymous_4276 Yes!
That thumbnail don't do that again
1
pretty good for being bad at these
open answer: "a symmetric inequality THE EGYPTIANS LEFT US or FROM THE DAWN OF TIME or KING EDWARD THE THIRD WAS VERY FOND OF or FOUND IN THE CORNER OF THE HOLY SHROUD OF TURIN or ALLOWING DOGS AND CATS TO LIVE TOGETHER WITHOUT THE MASS HYSTERIA (CIT.)"
Stop putting so many words in caps.
I would title it "You won't believe your eyes when you see this insane inequality!!". All in caps of course.
No, all caps is rude yelling.