Hello /u/---Kino---,
Your post has been removed because of this rule:
Your post is a math problem without any context. You are required to explain your attempts at solving the question, or specifically what you need help with. When you post again, please include a comment with more details such as listing the steps **you** have already tried, to solve the problem.
In the past, I've been downvoted to hell and back at times for suggesting even smaller changes - I just want to be as discrete and as polite as possible.
Yeah, people spray downvotes in maths forums for some reason. I've seen people get pounded for asking the sort of follow up questions you'd want a curious student to ask.
I'll be quite honest with you - and by the way thanks for supporting my sentiment! - I try to be helpful as much as the next guy, but discussing Mathematics works best face to face and next to a blackboard with a piece of chalk in hand - it's extremely difficult to drive a point across and explain ideas when all we're doing here is akin to turn based combat in games. Best of luck and happy redditing!
Haha, true.
I started doing this as a "pay it forward" for everyone on the internet who helped me get from "basic high school understanding" to being able to do more interesting things in my 30s.
I had no access to a teacher or classroom so it was quite hard, and would have been almost impossible if not for places like these.
I'm glad it actually helps people. Another driver for my activities here is the diversity of problems that one can find in this subreddit!
I used to collect books on mathematical puzzles, competitions, different mathematical subjects and I can say that this subreddit offers a comparable breadth of topics, and is all the more challenging because no two problems are even expected to be connected in any way.
Then use [Cauchy-Schwarz](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality) with u = (a,b,c) and v = (b,c,a). You should find a=b=c=4 achieves the minimum.
Mathematically, it’s valid inquiry. With a little bit of social context, including the part where OP posted about knowing that the numbers should all be the same, it’s pretty clear he was asking about the minimum and had made a typo.
I think this is a dramatic reenactment of the difference between programmer and product manager :)
If you don't want to apply Cauchy-Schwarz, consider the expression
a^2 + b^2 + c^2 -ab -ac -bc = a^2 - ab + b^2 - bc + c^2 - ac
and complete the squares. What you get is
a^2 + b^2 + c^2 -ab -ac -bc =
(a - b)^2 + ab - b^2 + (b - c)^2 + bc - c^2 + (c - a)^2 + ac - a^2.
Therefore
a^2 + b^2 + c^2 -ab -ac -bc = (a - b)^2 + ab - b^2 + (b - c)^2 + bc - c^2 + (c - a)^2 + ac - a^2
which is the same as
2(a^2 + b^2 + c^2 -ab -ac -bc) = (a - b)^2 + (b - c)^2 + (c - a)^2.
Since the RHS is a sum of squares, it is >=0, so 2(a^2 + b^2 + c^2 -ab -ac -bc)>=0 or equivalently a^2 + b^2 + c^2 >= ab + ac + bc.
Therefore the minimum value is 48. To find where the minimum is attained, note that (a - b)^2 + (b - c)^2 + (c - a)^2 attains its minimum at 0, and this is only possible when all three squares are 0, so a=b, b=c, c=a. Hence 3a^2 = 48 and a = 4 or -4, so the minimum is attained at (4,4,4) and (-4, -4, -4).
I just saw a question like this on the AMC 10 using similar algebraic manipulation https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_13
A (maybe) interesting geometric interpretation of this question:
ab + ac + bc is half of the surface area of a cuboid with sidelengths a, b and c.
a\^2 + b\^2 + c\^2 is the square of the length of the diagonal form one corner to the opposite corner.
So effectively the question is how to maximize the diagonal in a cuboid of given surface area.
Now consider a cuboid with very small side lengths a and b (effectively a very long rod). Its length c can grow above all limits while keeping the surface area constant (here to be 96).
But the length of the diagonal goes to c, when a and b tend to zero. So it also grows above all limits.
As the other commenter pointed out, the problem is likely asking for a lower bound (take a = 0, b= 48/c and any value of c works, which makes the second part arbitrarily large).
In that case, the easiest way I can think about is applying the Cauchy-Schwarz inequality (not sure if you are familiar with it). I will leave the rest of the work to you.
[Link to Wikipedia article](https://en.m.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality)
L = a^2 + b^2 + c^2 + x (48 - ab - ac - bc)
FOC with respect to a = 0
2a = xb + xc
FOC with respect to b = 0
2b = xa + xc
FOC with respect to c = 0
2c = xa + xb
FOC with respect to x = 0
48 = ab + ac + bc
Then you have four equations and four unknowns.
By symmetry, I think it’s pretty obvious a = b = c
So 48 = 3a^2
Thus a = 4 ( or negative 4)
And b and c also equal 4
Edit: I saw from another comment that you wanted minimum and not maximum.
Edit 2: I guess I didn’t really do shit to explain why all variables are equal.
This is a common problem that is solved in economics, so I view this problem in terms of utility and cost. You have a cost function (a^2 + b^2 + c^2) which you want to minimize subject to having a utility equal to 48. In order to “purchase” utility you need to incur cost. You therefore are comparing the marginal cost to the marginal utility associated with increasing each variable.
The marginal cost of each variable (2a for a, 2b for b and 2c for c) increases as that variable increases. Moreover, as one variable increases, the marginal utility of all other variables gets higher (for example, the marginal utility of a is b + c, so as b or c increases a becomes a more attractive purchase).
Because of this, you need to think at the margin - if one variable was greater than any of the others, you would get more utility and incur less costs if you were to swap that variable for one of the other variables. This necessarily means that all variables will be equal to each other.
Happy to explain better if this explanation was helpful.
Why is everyone saying that Cauchy would be the easiest way to solve it?
One could just say:
“By AM-GM we get
a^2 + b^2 greater than or equal to 2ab.
Similarly, we do this for bc and ca to get
2 (a^2 + b^2 + c^2) greater than or equal to 2 (ab + bc + ca)
So a^2 + b^2 + c^2 greater than or equal to ab + bc + ca = 48. Thus the minimum value is 48.”
Additionally, one could simply cite Muirhead.
“Since (2,0) majorizes (1,1), we have:
a^2 + b^2 + c^2 greater than or equal to ab + bc + ca = 48. Thus the minimum value is 48.”
Yep, absolutely agreed, this is a way nicer solution.
A third more elementary way would be using the Rearrangement Inequality. Wlog. a ≥ b ≥ c, so a²+b²+c² ≥ ab+ac+bc. Obviously equality is possible if a=b=c, so 48 is the minimum.
0 ≤ (a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2 + b^2 + c^2 - ab - ac - bc), hence
a^2 + b^2 + c^2 ≥ ab + ac + bc = 48
so the minimum is 48. There is no maximum, as seen by taking a arbitrary, b=48/a, c=0.
For a more advanced approach you could use the ‘Lagrangian multiplier’ method from multivariable calc. It’s a bit more involved however it is very procedural just like optimisation in single variable.
Let
a = 0
b = a real fuckton
c = 48 / a real fuckton, which is also real
Then
ac + ab + bc = 48
a^2 + b^2 + c^2 ~= b^2 = a real fuckton squared
You probably meant natural, not real.
Seems like no one mentioned it, but this is also a trivial application of the rearrangement inequality. Intuitively, it just says that you would want 5 $100 bills and 2 $10 bills instead of the other way around.
Given that a, b, and c are equally represented in both of the given expressions, we can assume that the ideal value will be a = b = c, therefore 3x\^2 = 48, so x = 4. Therefore, a = b = c = 4 is the minimum value.
Basicly the first equation means if you increase 1 variable you need to decrease the rest.
The way the 2nd equation works is you want to maximise 1 variable and make 0 the rest.
However, at least 2 numbers need to be non 0 for the first equation to be valid.
So i would go with 1 variable is 1 and the other is 48.
That should be the maximum.
Let f(x) = (x-a)(x-b)(x-c).
f(x) = x³ - Ax² + 48x - C
a² + b² + c² = A² - 96
f(x) must have three real roots (given a,b,c ∈ ℝ)
f'(x) needs to have two real roots
f'(x) = 3x² - 2Ax + 48
∆ = 4A² - 576 = 4(A² - 144) = 4(A-12)(A+12) > 0
Meaning A ∈ (-∞,-12) ∪ (12,+∞)
Furthermore, for these roots of f'(x), call them x_1 and x_2, we must have f(x_1)f(x_2) < 0
You could probably spend some time working out all the algebra here and get to some other restriction for A. And then argue that a²+b²+c² achieves its maximum wherever |A| achieves its maximum
Basically, realise that 0<=(a-b)^2 = a^2 + b^2 -2ab. From this you can see 1/2(a^2 +b^2 )>=ab, with equality only if a=b.
We therefore have: a^2 +b^2 +c^2 =1/2(a^2 +b^2 )+1/2(a^2 +c^2 )+1/2(b^2 +c^2 )>=ab+ac+bc=48, with equality only when a=b=c. This gives a minimum of 48, achieved when a=b=c=4.
It would be. OP already clarified in another comment it was meant to be a minimum.
The expression does not have a maximum. Take for instance a=48/x, b=x, c=0, with x an arbitrary constant. By making x big (or small), we can get a^2+b^2+c^2 to be arbitrarily large.
Hello /u/---Kino---, Your post has been removed because of this rule: Your post is a math problem without any context. You are required to explain your attempts at solving the question, or specifically what you need help with. When you post again, please include a comment with more details such as listing the steps **you** have already tried, to solve the problem.
Sorry if I am introducing unnecessary confusion, but are we sure we're looking for a maximum, not a minimum?
This is valid inquiry because there is no maximum. Just take (a,b,c)=(a,48/a,0) and make a as large as you like.
In the past, I've been downvoted to hell and back at times for suggesting even smaller changes - I just want to be as discrete and as polite as possible.
Yeah, people spray downvotes in maths forums for some reason. I've seen people get pounded for asking the sort of follow up questions you'd want a curious student to ask.
I'll be quite honest with you - and by the way thanks for supporting my sentiment! - I try to be helpful as much as the next guy, but discussing Mathematics works best face to face and next to a blackboard with a piece of chalk in hand - it's extremely difficult to drive a point across and explain ideas when all we're doing here is akin to turn based combat in games. Best of luck and happy redditing!
Haha, true. I started doing this as a "pay it forward" for everyone on the internet who helped me get from "basic high school understanding" to being able to do more interesting things in my 30s. I had no access to a teacher or classroom so it was quite hard, and would have been almost impossible if not for places like these.
I'm glad it actually helps people. Another driver for my activities here is the diversity of problems that one can find in this subreddit! I used to collect books on mathematical puzzles, competitions, different mathematical subjects and I can say that this subreddit offers a comparable breadth of topics, and is all the more challenging because no two problems are even expected to be connected in any way.
Yes im sorry its minimum
Then use [Cauchy-Schwarz](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality) with u = (a,b,c) and v = (b,c,a). You should find a=b=c=4 achieves the minimum.
Nice. Never seen it this way for sums of cross terms. I would straight go to either the symmetric solution or, to make sure, Lagrange multipliers.
My first instinct was Muirhead, but I decided it was a little overkill if it could be done by simpler inequalities like C-S.
Oh i never taught about that thanks
Mathematically, it’s valid inquiry. With a little bit of social context, including the part where OP posted about knowing that the numbers should all be the same, it’s pretty clear he was asking about the minimum and had made a typo. I think this is a dramatic reenactment of the difference between programmer and product manager :)
If you don't want to apply Cauchy-Schwarz, consider the expression a^2 + b^2 + c^2 -ab -ac -bc = a^2 - ab + b^2 - bc + c^2 - ac and complete the squares. What you get is a^2 + b^2 + c^2 -ab -ac -bc = (a - b)^2 + ab - b^2 + (b - c)^2 + bc - c^2 + (c - a)^2 + ac - a^2. Therefore a^2 + b^2 + c^2 -ab -ac -bc = (a - b)^2 + ab - b^2 + (b - c)^2 + bc - c^2 + (c - a)^2 + ac - a^2 which is the same as 2(a^2 + b^2 + c^2 -ab -ac -bc) = (a - b)^2 + (b - c)^2 + (c - a)^2. Since the RHS is a sum of squares, it is >=0, so 2(a^2 + b^2 + c^2 -ab -ac -bc)>=0 or equivalently a^2 + b^2 + c^2 >= ab + ac + bc. Therefore the minimum value is 48. To find where the minimum is attained, note that (a - b)^2 + (b - c)^2 + (c - a)^2 attains its minimum at 0, and this is only possible when all three squares are 0, so a=b, b=c, c=a. Hence 3a^2 = 48 and a = 4 or -4, so the minimum is attained at (4,4,4) and (-4, -4, -4).
Beautifully done while avoiding advanced concepts - nice
I just saw a question like this on the AMC 10 using similar algebraic manipulation https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_13
A (maybe) interesting geometric interpretation of this question: ab + ac + bc is half of the surface area of a cuboid with sidelengths a, b and c. a\^2 + b\^2 + c\^2 is the square of the length of the diagonal form one corner to the opposite corner. So effectively the question is how to maximize the diagonal in a cuboid of given surface area. Now consider a cuboid with very small side lengths a and b (effectively a very long rod). Its length c can grow above all limits while keeping the surface area constant (here to be 96). But the length of the diagonal goes to c, when a and b tend to zero. So it also grows above all limits.
really good solution but i wrote thw question wrong i should write minimum instead of maximum sorry
As the other commenter pointed out, the problem is likely asking for a lower bound (take a = 0, b= 48/c and any value of c works, which makes the second part arbitrarily large). In that case, the easiest way I can think about is applying the Cauchy-Schwarz inequality (not sure if you are familiar with it). I will leave the rest of the work to you. [Link to Wikipedia article](https://en.m.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality)
Oh wow, just saw the other comment.
L = a^2 + b^2 + c^2 + x (48 - ab - ac - bc) FOC with respect to a = 0 2a = xb + xc FOC with respect to b = 0 2b = xa + xc FOC with respect to c = 0 2c = xa + xb FOC with respect to x = 0 48 = ab + ac + bc Then you have four equations and four unknowns. By symmetry, I think it’s pretty obvious a = b = c So 48 = 3a^2 Thus a = 4 ( or negative 4) And b and c also equal 4 Edit: I saw from another comment that you wanted minimum and not maximum. Edit 2: I guess I didn’t really do shit to explain why all variables are equal. This is a common problem that is solved in economics, so I view this problem in terms of utility and cost. You have a cost function (a^2 + b^2 + c^2) which you want to minimize subject to having a utility equal to 48. In order to “purchase” utility you need to incur cost. You therefore are comparing the marginal cost to the marginal utility associated with increasing each variable. The marginal cost of each variable (2a for a, 2b for b and 2c for c) increases as that variable increases. Moreover, as one variable increases, the marginal utility of all other variables gets higher (for example, the marginal utility of a is b + c, so as b or c increases a becomes a more attractive purchase). Because of this, you need to think at the margin - if one variable was greater than any of the others, you would get more utility and incur less costs if you were to swap that variable for one of the other variables. This necessarily means that all variables will be equal to each other. Happy to explain better if this explanation was helpful.
Why is everyone saying that Cauchy would be the easiest way to solve it? One could just say: “By AM-GM we get a^2 + b^2 greater than or equal to 2ab. Similarly, we do this for bc and ca to get 2 (a^2 + b^2 + c^2) greater than or equal to 2 (ab + bc + ca) So a^2 + b^2 + c^2 greater than or equal to ab + bc + ca = 48. Thus the minimum value is 48.” Additionally, one could simply cite Muirhead. “Since (2,0) majorizes (1,1), we have: a^2 + b^2 + c^2 greater than or equal to ab + bc + ca = 48. Thus the minimum value is 48.”
nice solution thank
Yep, absolutely agreed, this is a way nicer solution. A third more elementary way would be using the Rearrangement Inequality. Wlog. a ≥ b ≥ c, so a²+b²+c² ≥ ab+ac+bc. Obviously equality is possible if a=b=c, so 48 is the minimum.
0 ≤ (a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2 + b^2 + c^2 - ab - ac - bc), hence a^2 + b^2 + c^2 ≥ ab + ac + bc = 48 so the minimum is 48. There is no maximum, as seen by taking a arbitrary, b=48/a, c=0.
Also, consider the effects of negatives.
For a more advanced approach you could use the ‘Lagrangian multiplier’ method from multivariable calc. It’s a bit more involved however it is very procedural just like optimisation in single variable.
Im too lazy to calculate it but i would use the Karush Kuhn Tucker method for calculating minimums from non-linear programmimg
Let a = 0 b = a real fuckton c = 48 / a real fuckton, which is also real Then ac + ab + bc = 48 a^2 + b^2 + c^2 ~= b^2 = a real fuckton squared You probably meant natural, not real.
Seems like no one mentioned it, but this is also a trivial application of the rearrangement inequality. Intuitively, it just says that you would want 5 $100 bills and 2 $10 bills instead of the other way around.
Given that a, b, and c are equally represented in both of the given expressions, we can assume that the ideal value will be a = b = c, therefore 3x\^2 = 48, so x = 4. Therefore, a = b = c = 4 is the minimum value.
Basicly the first equation means if you increase 1 variable you need to decrease the rest. The way the 2nd equation works is you want to maximise 1 variable and make 0 the rest. However, at least 2 numbers need to be non 0 for the first equation to be valid. So i would go with 1 variable is 1 and the other is 48. That should be the maximum.
i can do this only with lagrange theorem if you are interested in that solution
Let f(x) = (x-a)(x-b)(x-c). f(x) = x³ - Ax² + 48x - C a² + b² + c² = A² - 96 f(x) must have three real roots (given a,b,c ∈ ℝ) f'(x) needs to have two real roots f'(x) = 3x² - 2Ax + 48 ∆ = 4A² - 576 = 4(A² - 144) = 4(A-12)(A+12) > 0 Meaning A ∈ (-∞,-12) ∪ (12,+∞) Furthermore, for these roots of f'(x), call them x_1 and x_2, we must have f(x_1)f(x_2) < 0 You could probably spend some time working out all the algebra here and get to some other restriction for A. And then argue that a²+b²+c² achieves its maximum wherever |A| achieves its maximum
Basically, realise that 0<=(a-b)^2 = a^2 + b^2 -2ab. From this you can see 1/2(a^2 +b^2 )>=ab, with equality only if a=b. We therefore have: a^2 +b^2 +c^2 =1/2(a^2 +b^2 )+1/2(a^2 +c^2 )+1/2(b^2 +c^2 )>=ab+ac+bc=48, with equality only when a=b=c. This gives a minimum of 48, achieved when a=b=c=4.
I do apologize, but wouldn't that be 48 for a minimum?
It would be. OP already clarified in another comment it was meant to be a minimum. The expression does not have a maximum. Take for instance a=48/x, b=x, c=0, with x an arbitrary constant. By making x big (or small), we can get a^2+b^2+c^2 to be arbitrarily large.
Excellent, thank you - but in your original comment what do you mean by "This gives a minimum of 96, achieved when a=b=c=4."?
Oops, you're completely right!
ab<=(a^2 +b^2 )/2 bc<=(b^2 +c^2 )/2 ac<=(a^2 +c^2 )/2 48=ab+bc+ac<=(a^2 +b^2 )/2+(b^2 +c^2 )/2+(a^2 +c^2 )/2=a^2 +b^2 +c^2 a^2 +b^2 +c^2 >=48 The equality works if a=b=c=4
Think what (a+b+c)\^2 looks like.