# Math Challenge - December 2019

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13. There is only one rule at Christmas at the world's richest family: The gifts have to be expensive, heavy and glamorous. So they all present statues of pure gold. It may be large figure, a tiger sculpture or an opulent candlestick. The eldest son who doesn't live at home anymore receives gifts of nine tons total, but none of which is heavier than a ton. He wants to bring home all of them, but only could rent trucks which can load three tons maximal. How many trucks are needed to at least be able to transport all gifts of gold at the same time?

Shouldn't the question be asking what is the maximum but optimal number of trucks needed rather than the minimum? The minimum must be 3 because with fewer then 3 trucks, it is not possible to transport 9 tons in one go since then the trucks will be required to carry more than 3 tons which is beyond their limit. For example, if there were a total 9 gifts each weighing 1 ton, then 3 trucks would be necessary but also sufficient as each of the 3 trucks can be loaded with 3 items each, so each truck carries 3 tons.

Answer: 4 trucks are always sufficient to transport all items at the same time, whatever be the number of gift items and their individual weights provided no one of them exceeds 1 ton in weight and they add up to 9 tons.

First, we show that 3 trucks are not always sufficient using a counterexample. If the gifts consisted of 4 items weighing 0.975 tons each, 3 items weighing 0.9 tons each and 3 items weighing 0.8 tons each, then we have 10 items with a total weight of 9 tons. In this case, no single truck can carry more than 3 items in one go because the minimum weight per item is 0.8 tons and loading 4 items would mean exceeding the 3 ton limit. So the 3 trucks together cannot carry more than 9 items at the same time. Since the total number of items is 10, it is proven that 3 trucks are insufficient.

It follows that the final total unused capacity of the 3 trucks with the above loading strategy, ##f_{total} = f_1 + f_2 + f_3 \lt 3##. The total used capacity of the 3 trucks would therefore be ##u_{total} = (3 - f_{1}) + (3 - f_{2}) + (3 - f_{3}) = 9 - f_{total} \gt 6##. Since the total weight of all gifts if 9 tons, and more than 6 tons could be accommodated in 3 trucks as per the above equations, it follows that the total weight of the remaining "unaccommodated" items must be less than 3 tons. And all these remaining items can be loaded in a 4th truck of carrying capacity 3 tons.

No, the one dimensional case is a) not meant and b) different from what you wrote.

The one dimensional "square" of side length ##4## is ##[0,4]##. The "circles" in the corners of radius ##1## would be ##[0,2]## and ##[2,4]##, which leaves the point ##\{\,2\,\}## as the central "circle", which does not touch the boundary ##\{\,0,4\,\}## of the "cube" ##[0,4]## at all.

The trick is to calculate the radius of the inner hyperspheres. This radius is ##0## in one dimension.

Thanks for reviewing that attempted solution and clarifying. I am glad that I had at least got the concepts of 1-D "circles" and 1-D "central circle" correct in that attempt. My understanding of the central circle touching the boundary of the "square"/"cube" was that it had to touch one point on each "edge"/"face" of the "square"/"cube". As per that interpretation, the 1-D "central circle", i.e. the mid-point in my solution, does touch the boundary in the 1-D case. By boundaries, does the question mean the corner points?

"The trick is to calculate the radius of the inner hyperspheres." - I guessed the same initially but wanted to validate the 1-D case 1st

Mentor
Shouldn't the question be asking what is the maximum but optimal number of trucks needed rather than the minimum? The minimum must be 3 because with fewer then 3 trucks, it is not possible to transport 9 tons in one go since then the trucks will be required to carry more than 3 tons which is beyond their limit. For example, if there were a total 9 gifts each weighing 1 ton, then 3 trucks would be necessary but also sufficient as each of the 3 trucks can be loaded with 3 items each, so each truck carries 3 tons.
This is hair splitting. Every number of insufficiently many trucks is a minimum as every number larger than necessarily many trucks is a maximum. The exact wording would be: The minimal number of necessary trucks, the least upper bound, the supremum of all numbers of sufficiently many trucks. The supremum is a minimum, as it is the least upper bound, and it is a maximum, as it is an upper bound of necessary trucks. Hence the new word. The greatest lower bound would called infimum. So we could likewise say, that we are looking for the infimum of all numbers of insufficiently many trucks plus one.
Answer: 4 trucks are always sufficient to transport all items at the same time, whatever be the number of gift items and their individual weights provided no one of them exceeds 1 ton in weight and they add up to 9 tons.

First, we show that 3 trucks are not always sufficient using a counterexample. If the gifts consisted of 4 items weighing 0.975 tons each, 3 items weighing 0.9 tons each and 3 items weighing 0.8 tons each, then we have 10 items with a total weight of 9 tons. In this case, no single truck can carry more than 3 items in one go because the minimum weight per item is 0.8 tons and loading 4 items would mean exceeding the 3 ton limit. So the 3 trucks together cannot carry more than 9 items at the same time. Since the total number of items is 10, it is proven that 3 trucks are insufficient.
Right. Easier would have been: ##10## gifts ##900\,kg## each.

It follows that the final total unused capacity of the 3 trucks with the above loading strategy, ##f_{total} = f_1 + f_2 + f_3 \lt 3##. The total used capacity of the 3 trucks would therefore be ##u_{total} = (3 - f_{1}) + (3 - f_{2}) + (3 - f_{3}) = 9 - f_{total} \gt 6##. Since the total weight of all gifts if 9 tons, and more than 6 tons could be accommodated in 3 trucks as per the above equations, it follows that the total weight of the remaining "unaccommodated" items must be less than 3 tons. And all these remaining items can be loaded in a 4th truck of carrying capacity 3 tons.
Well done, albeit it bit complicated. You should study math! Mathematicians tend to either a totally trivial solution or a complicated one!

So I invite everyone to find an easier argument, why 4 trucks are sufficient!

Problem 7. I don't know how to express, in closed form, the double sum,$$\sum_{n=0}^{\infty}\sum_{k=0}^n (\frac{3}{2})^k \frac{(2n-2k)!(2k)!}{8^n \left [(n-k)!\right ]^2 \left [(k)!\right ]^2 \left (2n(1+2k \right )+ (1-4k^2))}$$ However, for anyone that may be interested, the double sum,$$\sum_{n=0}^{\infty}\sum_{k=0}^n (\frac{3}{2})^k \frac{(2n-2k)!(2k)!}{8^n \left [(n-k)!\right ]^2 \left [(k)!\right ]^2 \left (2n(1+2k \right )- (1+4k^2))}$$can be expressed in closed form as follows. I note that the double sum can be expressed as ,$$\sum_{n=0}^{\infty}\sum_{k=0}^na_{k,n-k}=\sum_{m=0}^{\infty}\sum_{k=0}^{\infty} a_{k,m} \\ m=n-k$$where ##a_{k,m}## are terms in the expansion. On making the substitution ##m=n-k## in the double sum,$$\sum_{n=0}^{\infty}\sum_{k=0}^n (\frac{3}{2})^k \frac{(2n-2k)!(2k)!}{8^n \left [(n-k)!\right ]^2 \left [(k)!\right ]^2 \left (2n(1+2k \right )- (1+4k^2))} \\ =\sum_{m=0}^{\infty}\sum_{k=0}^{\infty} (\frac{3}{2})^{k} \frac{2(m)!(2k)!}{8^{m+k} \left [(m)!\right ]^2 \left [(k)!\right ]^2 \left (2(m+k)(1+2k \right )- (1+4k^2))} \\ =\sum_{m=0}^{\infty} \sum_{k=0}^{\infty} (\frac{1}{2})^{m} (\frac{3}{4})^{k} \frac{(2m)!(2k)!}{4^m 4^k\left [(m)!\right ]^2 \left [(k)!\right ]^2 (2m+1)(2k+1)} \\ =2\sqrt{\frac{2}{3}}\sum_{m=0}^{\infty} \sum_{k=0}^{\infty} (\frac{1}{\sqrt{2}})^{2m+1} (\frac{\sqrt{3}}{2})^{2k+1} \frac{(2m)!(2k)!}{4^m 4^k\left [(m)!\right ]^2 \left [(k)!\right ]^2 (2m+1)(2k+1)} \\=2\sqrt{\frac{2}{3}} \sum_{m=0}^{\infty}(\frac{1}{\sqrt{2}})^{2m+1}\frac{(2m)!}{4^m [(m)! ]^2 (2m+1)} \sum_{k=0}^{\infty}(\frac{\sqrt{3}}{2} )^{2k+1}\frac{(2k)!}{4^k[(k)! ]^2 (2k+1)}$$I expand the function##\frac{1} {\sqrt{1-x^2}}## in a binomial series,$$\frac{1} {\sqrt{1-x^{2}}}=\sum_{n=0}^{\infty}\begin{pmatrix} \frac{-1}{2} \\ n \end{pmatrix} x^{2n} =\sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n!)^2}x^{2n}$$and I recall that,$$\sin^{-1}(x)=\int_0^x \frac{dt}{\sqrt{1-t^2}} \\=\sum_{n=0}^{\infty} \int_0^x\frac{(2n)!}{4^n(n!)^2}x^{2n}dx \\=\sum_{n=0}^{\infty} \frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n +1}$$On comparison with the double sum I deduce,$$\sum_{n=0}^{\infty}\sum_{k=0}^n (\frac{3}{2})^k \frac{(2n-2k)!(2k)!}{8^n \left [(n-k)!\right ]^2 \left [(k)!\right ]^2 \left (2n(1+2k \right )- (1+4k^2))} \\=2\sqrt{\frac{2}{3}}\sin^{-1} (\frac{1}{\sqrt{2}})\sin^{-1} (\frac{\sqrt{3}}{2})$$

Mentor
Thanks for reviewing that attempted solution and clarifying. I am glad that I had at least got the concepts of 1-D "circles" and 1-D "central circle" correct in that attempt. My understanding of the central circle touching the boundary of the "square"/"cube" was that it had to touch one point on each "edge"/"face" of the "square"/"cube".
Yes. The symmetry of the problem however, guarantees that one point on one edge and one point on every edge (which are not vertices, hence necessarily pairwise different) are equivalent.

The central circle is defined by
In the center of the square is another circle that touches the other four.
The other four what? Since circle is the only object available in this sentence, it is allowed to refer to it without repetition. Such a sentence has always to be read as "In the center of the square is another circle that touches the other four [circles]." A specification is only necessary in case more than one object is available as a possible reference. But there are only four circles of that kind. The other object which has four somethings are the edges, which are a) no circles and b) not been mentioned in the other part of the sentence, and can thus be ruled out.

However, we don't want to study grammar here - the more as English isn't my native language. So in our context it is sufficient to simply ask!
As per that interpretation, the 1-D "central circle", i.e. the mid-point in my solution, does touch the boundary in the 1-D case. By boundaries, does the question mean the corner points?
No, and no. ##\{\,2\,\}## touches the boundaries of ##[0,2]## and ##[2,4]##, hence it is the central circle. But it does not touch the boundary (edges, hyperfaces) of the square (hypercube) which are ##\{\,0\,\}## and ##\{\,4\,\}## in this trivial one dimensional case.
"The trick is to calculate the radius of the inner hyperspheres." - I guessed the same initially but wanted to validate the 1-D case 1st
This is not really a case because all objects become intervals and boundaries single points.

Do not force me to define the entire thing mathematically correct in ##\mathbb{R}^n##. It is sill a high school question which can be solved with high school methods. A mathematical rigorous description would exceed the length of the argument to prove the statement! At least if I had to use high school language.

Mentor
Problem 7. I don't know how to express, in closed form, the double sum,$$\sum_{n=0}^{\infty}\sum_{k=0}^n (\frac{3}{2})^k \frac{(2n-2k)!(2k)!}{8^n \left [(n-k)!\right ]^2 \left [(k)!\right ]^2 \left (2n(1+2k \right )+ (1-4k^2))}$$ However, for anyone that may be interested, the double sum,$$\sum_{n=0}^{\infty}\sum_{k=0}^n (\frac{3}{2})^k \frac{(2n-2k)!(2k)!}{8^n \left [(n-k)!\right ]^2 \left [(k)!\right ]^2 \left (2n(1+2k \right )- (1+4k^2))}$$can be expressed in closed form as follows. I note that the double sum can be expressed as ,$$\sum_{n=0}^{\infty}\sum_{k=0}^na_{k,n-k}=\sum_{m=0}^{\infty}\sum_{k=0}^{\infty} a_{k,m} \\ m=n-k$$where ##a_{k,m}## are terms in the expansion. On making the substitution ##m=n-k## in the double sum,$$\sum_{n=0}^{\infty}\sum_{k=0}^n (\frac{3}{2})^k \frac{(2n-2k)!(2k)!}{8^n \left [(n-k)!\right ]^2 \left [(k)!\right ]^2 \left (2n(1+2k \right )- (1+4k^2))} \\ =\sum_{m=0}^{\infty}\sum_{k=0}^{\infty} (\frac{3}{2})^{k} \frac{2(m)!(2k)!}{8^{m+k} \left [(m)!\right ]^2 \left [(k)!\right ]^2 \left (2(m+k)(1+2k \right )- (1+4k^2))} \\ =\sum_{m=0}^{\infty} \sum_{k=0}^{\infty} (\frac{1}{2})^{m} (\frac{3}{4})^{k} \frac{(2m)!(2k)!}{4^m 4^k\left [(m)!\right ]^2 \left [(k)!\right ]^2 (2m+1)(2k+1)} \\ =2\sqrt{\frac{2}{3}}\sum_{m=0}^{\infty} \sum_{k=0}^{\infty} (\frac{1}{\sqrt{2}})^{2m+1} (\frac{\sqrt{3}}{2})^{2k+1} \frac{(2m)!(2k)!}{4^m 4^k\left [(m)!\right ]^2 \left [(k)!\right ]^2 (2m+1)(2k+1)} \\=2\sqrt{\frac{2}{3}} \sum_{m=0}^{\infty}(\frac{1}{\sqrt{2}})^{2m+1}\frac{(2m)!}{4^m [(m)! ]^2 (2m+1)} \sum_{k=0}^{\infty}(\frac{\sqrt{3}}{2} )^{2k+1}\frac{(2k)!}{4^k[(k)! ]^2 (2k+1)}$$I expand the function##\frac{1} {\sqrt{1-x^2}}## in a binomial series,$$\frac{1} {\sqrt{1-x^{2}}}=\sum_{n=0}^{\infty}\begin{pmatrix} \frac{-1}{2} \\ n \end{pmatrix} x^{2n} =\sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n!)^2}x^{2n}$$and I recall that,$$\sin^{-1}(x)=\int_0^x \frac{dt}{\sqrt{1-t^2}} \\=\sum_{n=0}^{\infty} \int_0^x\frac{(2n)!}{4^n(n!)^2}x^{2n}dx \\=\sum_{n=0}^{\infty} \frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n +1}$$On comparison with the double sum I deduce,$$\sum_{n=0}^{\infty}\sum_{k=0}^n (\frac{3}{2})^k \frac{(2n-2k)!(2k)!}{8^n \left [(n-k)!\right ]^2 \left [(k)!\right ]^2 \left (2n(1+2k \right )- (1+4k^2))} \\=2\sqrt{\frac{2}{3}}\sin^{-1} (\frac{1}{\sqrt{2}})\sin^{-1} (\frac{\sqrt{3}}{2})$$
Very good! Which number is that?

Mentor
@fresh_42 Problem 7
No, I'm asking for the actual result. Which number is your expression?

No, I'm asking for the actual result. Which number is your expression?
Silly me. ##\sin^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}## and ##\sin^{-1}(\frac{\sqrt{3}}{2})=\cos^{-1}(\frac{1}{2})=\frac{\pi}{3}## so I get ##\frac{1}{6} \sqrt{\frac{2}{3}} \pi##.
Thank you for posting interesting and difficult problems and commenting on their solutions. It reflects the breadth of your knowledge and patience, and I admire you for that.

Mentor
Silly me. ##\sin^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}## and ##\sin^{-1}(\frac{\sqrt{3}}{2})=\cos^{-1}(\frac{1}{2})=\frac{\pi}{3}## so I get ##\frac{1}{6} \sqrt{\frac{2}{3}} \pi##.
Thank you for posting interesting and difficult problems and commenting on their solutions. It reflects the breadth of your knowledge and patience, and I admire you for that.
Not silly, but not quite right (forgotten the square ):
$$2\sqrt{\frac{2}{3}}\sin^{-1} \left(\frac{1}{\sqrt{2}}\right)\sin^{-1} \left(\frac{\sqrt{3}}{2}\right)=2\sqrt{\frac{2}{3}}\dfrac{\pi}{4}\dfrac{\pi}{3}= \dfrac{\pi^2}{6}\sqrt{\dfrac{2}{3}}=\dfrac{1}{3\sqrt{6}}\,\pi^2$$

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Not silly, but not quite right (forgotten the square):
$$2\sqrt{\frac{2}{3}}\sin^{-1} \left(\frac{1}{\sqrt{2}}\right)\sin^{-1} \left(\frac{\sqrt{3}}{2}\right)=2\sqrt{\frac{2}{3}}\dfrac{\pi}{4}\dfrac{\pi}{3}= \dfrac{\pi^2}{6}\sqrt{\dfrac{2}{3}}=\dfrac{1}{3\sqrt{6}}\,\pi^2$$
I guess my not being able to do 5th grade arithmetic is not silly but rather indicative of onset dementia.

fresh_42
I cannot follow you. In your very first equation I get different terms:
$$\sum_{k,j=^1}^\infty \dfrac{1}{kj(k+j)^2}=\underbrace{\dfrac{1}{1\cdot 2^2}}_{(1,1)}+\underbrace{\dfrac{2}{2\cdot 3^2}}_{(1,2)}+\underbrace{\dfrac{1}{4\cdot 4^2}}_{(2,2)}+\underbrace{\dfrac{2}{3\cdot 4^2}}_{(1,3)}+\underbrace{\dfrac{2}{6\cdot 5^2}}_{(2,3)}+\underbrace{\dfrac{1}{9\cdot 6^2}}_{(3,3)}+\ldots$$
so you must have used a summation I cannot see. And the result is a different one.

The trick is indeed to rewrite the summands. My proof uses integrals.
I don't follow what you don't follow. $$k=j=n \\ \sum_{k,j=1}^{\infty}\frac{1}{kj(k+j)^2}=\sum_{n=1}^{\infty}\frac{1}{n^2(2n)^2}=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^4}$$In my post on this problem I claim,$$\sum_{k,j=1}^{\infty}\frac{1}{kj(k+j)^2}=\frac{\pi^4}{360}$$I ran the following C code which iterates the sum 10000 times:

double z, zsum = 0, dz;
for (int i = 1; i < 10000; i++)
{
dz = (double)i;
z = 1.0 / pow(dz, 4);
zsum += z;
}
zsum *= 0.25;
The zsum variable agrees with ##\frac{\pi^4}{360}## computed on my HP calculator to 10 decimal places. QED I think. I request "solved by" honor.

Mentor
You cannot simply assume ##k=j##. Ok, you can, but this is not what a double sum means. See post #18 or the discussion before.

14. Prove ##\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq \dfrac{3}{2}## for ##a,b,c>0##

My solution uses basic calculus but I think there could be a simpler proof that doesn't need calculus.

Let ##X = \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}## and ##s = a + b+ c##.
##X## can be rewritten as ##X = \dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c} = \dfrac{A}{1-A}+\dfrac{B}{1-B}+\dfrac{C}{1-C}## where ##A = \dfrac{a}{s}, B = \dfrac{b}{s}, C = \dfrac{c}{s}##.

Since ##a, b, c > 0##, ##A, B, C \in (0, 1)## and ##A+B+C = 1##
Substituting ##C = 1 - A - B## in the earlier equation for ##X## gives ##X = \dfrac{A}{1-A}+\dfrac{B}{1-B}+\dfrac{1}{A+B} - 1##.

So ##X## is really a function of just 2 variables, ##A, B##. At the minimum value of ##X## for a fixed value of ##B##, we must have ##\frac {\partial X} {\partial A} = 0 \Rightarrow \frac {1} {1-A} + \frac {A} {(1-A)^2} - \frac {1} {(A+B)^2} = 0 \Rightarrow A=\frac {1-B} {2}## (there is no finite maximum since for any fixed B, ##A \rightarrow (1-B) \Rightarrow X \rightarrow \infty##, so the differential equaling 0 will correspond to a minimum)

Substituting for ##A## gives ##X_{min}(B)##, the minimum value of ##X## for a given value of ##B## to be ##\frac {3B^2 - 3B + 2} {1-B^2}##

To find the minimum of minimums, we set the differential of ##X_{min}## w.r.t. ##B## to 0.

##\frac {dX_{min}} {dB} = 0 \Rightarrow \frac {10B - 3B^2 - 3} {(1-B^2)^2} = 0 \Rightarrow 10B - 3B^2 - 3 = 0## (since the denominator cannot be infinite due to ##B \in (0, 1)##, equality to 0 needs numerator to be 0). The roots of this quadratic equation are ##B= 1/3## and ##B = 3## but only ##B = 1/3## is valid due to ##B \in (0, 1)##. So the minimum value of ##X## is achieved when ##B = 1/3## and ##X_{min}(1/3) = 3/2 \Rightarrow X \geq 3/2##.

Mentor
My solution uses basic calculus but I think there could be a simpler proof that doesn't need calculus.

Let ##X = \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}## and ##s = a + b+ c##.
##X## can be rewritten as ##X = \dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c} = \dfrac{A}{1-A}+\dfrac{B}{1-B}+\dfrac{C}{1-C}## where ##A = \dfrac{a}{s}, B = \dfrac{b}{s}, C = \dfrac{c}{s}##.

Since ##a, b, c > 0##, ##A, B, C \in (0, 1)## and ##A+B+C = 1##
Substituting ##C = 1 - A - B## in the earlier equation for ##X## gives ##X = \dfrac{A}{1-A}+\dfrac{B}{1-B}+\dfrac{1}{A+B} - 1##.

So ##X## is really a function of just 2 variables, ##A, B##. At the minimum value of ##X## for a fixed value of ##B##, we must have ##\frac {\partial X} {\partial A} = 0 \Rightarrow \frac {1} {1-A} + \frac {A} {(1-A)^2} - \frac {1} {(A+B)^2} = 0 \Rightarrow A=\frac {1-B} {2}## (there is no finite maximum since for any fixed B, ##A \rightarrow (1-B) \Rightarrow X \rightarrow \infty##, so the differential equaling 0 will correspond to a minimum)

Substituting for ##A## gives ##X_{min}(B)##, the minimum value of ##X## for a given value of ##B## to be ##\frac {3B^2 - 3B + 2} {1-B^2}##

To find the minimum of minimums, we set the differential of ##X_{min}## w.r.t. ##B## to 0.

##\frac {dX_{min}} {dB} = 0 \Rightarrow \frac {10B - 3B^2 - 3} {(1-B^2)^2} = 0 \Rightarrow 10B - 3B^2 - 3 = 0## (since the denominator cannot be infinite due to ##B \in (0, 1)##, equality to 0 needs numerator to be 0). The roots of this quadratic equation are ##B= 1/3## and ##B = 3## but only ##B = 1/3## is valid due to ##B \in (0, 1)##. So the minimum value of ##X## is achieved when ##B = 1/3## and ##X_{min}(1/3) = 3/2 \Rightarrow X \geq 3/2##.
Well, yes, there is an easier solution which uses the relation between certain means.
Your approach makes me wonder whether you're still in highschool?

Anyway, I could follow you except for the minimum argument. I do not see that ##\lim_{A\to 1-B} X = \infty ## since you have a fixed ##B## which fixes ##X=X(B)##, too.

However, we don't want to study grammar here - the more as English isn't my native language. So in our context it is sufficient to simply ask!

...
Do not force me to define the entire thing mathematically correct in ##\mathbb{R}^n##. It is sill a high school question which can be solved with high school methods. A mathematical rigorous description would exceed the length of the argument to prove the statement! At least if I had to use high school language.

Sorry if my reply gave the impression that your question was not clearly worded or was grammatically incorrect. Your question is fairly clear to me now. This is the first time that I am working on a problem involving geometry in more than 3 dimensions but as you have alluded to, the solution appears to need only an application of school-level maths. Will make another attempt at the solution during free time

Well, yes, there is an easier solution which uses the relation between certain means.

Yes, the form of the transformed equation and the presence of reciprocals in the denominators did bear some resemblance to some inequality relating to harmonic mean and simple mean but since I haven't studied or applied before identities relating to harmonic mean, only vaguely remembered reading those elsewhere, I didn't attempt solving using those. But now that you have mentioned, I looked on the web to verify the identity and was able to worked out proof based on it.

The original equation can be rewritten as ##X = \dfrac{S-P}{P} + \dfrac{S-Q}{Q} + \dfrac{S-R}{R} = \dfrac{S}{P} + \dfrac{S}{Q} + \dfrac{S}{R} - 3## where ##P=B+C, Q=A+C, R=A+B##

Since arithmetic mean >= harmonic mean, we get ##\dfrac {(\dfrac{S}{P} + \dfrac{S}{Q} + \dfrac{S}{R})} {3} \geq \dfrac {3} {\dfrac{P}{S} + \dfrac{Q}{S} + \dfrac{R}{S}}##. But since ##P+Q+R = 2S##, the right hand side of the inequality becomes ##3/2##. So we get ##\dfrac {(\dfrac{S}{P} + \dfrac{S}{Q} + \dfrac{S}{R})} {3} \geq 3/2 \Rightarrow (\dfrac{S}{P} + \dfrac{S}{Q} + \dfrac{S}{R}) \geq 9/2 \Rightarrow X \geq 9/2 - 3 = 3/2##

Anyway, I could follow you except for the minimum argument. I do not see that ##\lim_{A\to 1-B} X = \infty ## since you have a fixed ##B## which fixes ##X=X(B)##, too.

Sorry, that was a mistake on my part. I miscalculated the value of ##X## when ##A \to 1-B##. ##X## does not approach infinity if ##B## is fixed, but it should be possible to show that if ##A## moves away from ##(1-B)/2##, the value of ##X## increases, so ##A = \frac {1-B} {2}## must correspond to the minimum for a fixed ##B##. Will work out a proof if needed.

Mentor
Yes, the form of the transformed equation and the presence of reciprocals in the denominators did bear some resemblance to some inequality relating to harmonic mean and simple mean but since I haven't studied or applied before identities relating to harmonic mean, only vaguely remembered reading those elsewhere, I didn't attempt solving using those. But now that you have mentioned, I looked on the web to verify the identity and was able to worked out proof based on it.

The original equation can be rewritten as ##X = \dfrac{S-P}{P} + \dfrac{S-Q}{Q} + \dfrac{S-R}{R} = \dfrac{S}{P} + \dfrac{S}{Q} + \dfrac{S}{R} - 3## where ##P=B+C, Q=A+C, R=A+B##

Since arithmetic mean >= harmonic mean, we get ##\dfrac {(\dfrac{S}{P} + \dfrac{S}{Q} + \dfrac{S}{R})} {3} \geq \dfrac {3} {\dfrac{P}{S} + \dfrac{Q}{S} + \dfrac{R}{S}}##. But since ##P+Q+R = 2S##, the right hand side of the inequality becomes ##3/2##. So we get ##\dfrac {(\dfrac{S}{P} + \dfrac{S}{Q} + \dfrac{S}{R})} {3} \geq 3/2 \Rightarrow (\dfrac{S}{P} + \dfrac{S}{Q} + \dfrac{S}{R}) \geq 9/2 \Rightarrow X \geq 9/2 - 3 = 3/2##

Sorry, that was a mistake on my part. I miscalculated the value of ##X## when ##A \to 1-B##. ##X## does not approach infinity if ##B## is fixed, but it should be possible to show that if ##A## moves away from ##(1-B)/2##, the value of ##X## increases, so ##A = \frac {1-B} {2}## must correspond to the minimum for a fixed ##B##. Will work out a proof if needed.
Not really, as you now have proven if differently. My suspicion is that moving ##A## a little increases ##X##, too. I just wanted to emphasize that we have to be careful with applying one dimensional techniques on two dimensional cases. You considered ##B## fixed as you wrote ##\dfrac{\partial X}{\partial A}##, but ##B## isn't fixed. Maybe there is a global minimum elsewhere, and not the local ones on curves parametrized by ##B##. We need additional assumptions, e.g. that there are no isolated points. This is given here, but it has to be mentioned, that any minimum lies on such a differentiable curve with fixed ##B##. You implicitly used that the solution depends differentiable on ##A## and ##B##.

12. In a square of side length 4, there is a circle of radius 1 in each corner. In the center of the square is another circle that touches the other four. Analogously, in the three-dimensional case, in the center of a cube of edge length 4, there would be a sphere which would touch eight spheres of radius 1 placed in the corners of the cube. In which dimension does the central hypersphere become so large that it touches all sides of the hypercube?

Another attempt at solving this.
Is the answer 9? Here is how I arrived at the number. (Note: I have not attempted to prove the assumptions made in this solution, such as why the meeting point between the central hypersphere and and an outer hypersphere should lie on a particular line or why there would just one such meeting point per pair, partly because that will make the solution too verbose and partly because I am not familiar with higher-dimensional geometry to be able to give good formal proofs at first attempt)

As in my earlier attempt, I assume, for the sake of simplicity, but without loss of generality, that the square/cube has one of its corners coinciding with the origin and that every edge is lies on or is parallel to some axis. Suppose the number of axes (dimensions) is ##n##. Then the origin is given by the point ##(0, 0, .., 0)_n## in n-dimensional space, the center of the "bottom left" hypersphere (the one closest to the origin) is given by the n-dimensional point ##(1, 1, .., 1)_n## and the center of the central hypersphere is given by ##(2, 2, .., 2)_n##.

Intuitively, the meeting point between central hypersphere and the bottom left hypersphere should lie on the line segment connecting the 2 centers, i.e. on ##[(1,1,..,1)_n, (2,2,..,2)_n]## and therefore should be of the form ##(a,a,..,a)_n## (i.e. distance from origin is a same along all axes) and ##1 \lt a \leq 2##.

Now, the distance between ##[(1,1,..,1)_n## and ##(a,a,..,a)_n## should be the radius ##R## of the corner hyperspheres. ##R = \sqrt {n (a - 1)^2} = \sqrt {n} (a - 1)##. Since the radius ##R## of these is 1 as per the problem statement, we get ##a = 1 + \frac {1}{\sqrt {n}}##

Since the distance between ##(a,a,..,a)_n## and ##(2,2,..,2)_n## must be the radius ##r## of the central hypersphere, we get ##r = \sqrt {n (2 - a)^2} = \sqrt {n (2 - (1 + \frac {1}{\sqrt {n}}))^2} = \sqrt {n (1 - \frac {1}{\sqrt {n}})^2} = \sqrt {n} - 1##.

##r = 2## when ##n = 9## whereas for ##n \lt 9 \Rightarrow r \lt 2##. When the radius ##r## is exactly 2, the hypersphere centered at ##(2, 2, ..,2)_n## will touch the sides/faces of the hypercube (since the length of the cube is 4 on all axes) whereas with a smaller radius, the hypersphere will be contained entirely within the hypercube. Therefore, 9 is the dimension at which the condition stated in the question is met.

Mentor
When the radius rrr is exactly ##2##, the hypersphere centered at ##(2, 2, ..,2)_n## will touch the sides/faces of the hypercube (since the length of the cube is ##4## on all axes) whereas with a smaller radius, the hypersphere will be contained entirely within the hypercube. Therefore, ##9## is the dimension at which the condition stated in the question is met.
Well done! Yes, the radius of the center ball is ##\sqrt{n}-1## and ##9## the solution. In dimensions higher than ##9##, parts of the inner ball are even outside the box!

15. Let ##\mathbf{x}=(x_1,\ldots ,x_n)\, , \,\mathbf{y}=(y_1,\ldots ,y_n)## be tuples of positive numbers. Prove

##\prod_{k=1}^n(x_k+y_k)^{1/n} ≥ \prod_{k=1}^n {x_k}^{1/n} + \prod_{k=1}^n {y_k}^{1/n}##​

To prove the given inequality, we prove its equivalent (obtained by the dividing both sides of the original inequality by the LHS), ##1 ≥ \prod_{k=1}^n {\frac {x_k} {(x_k+y_k)}}^{1/n} + \prod_{k=1}^n {\frac {y_k} {(x_k+y_k)}}^{1/n}##. Since ##x_k, y_k## are positive numbers, we have ##0 \lt \frac {x_k} {(x_k+y_k)} \lt 1## and ##0 \lt \frac {y_k} {(x_k+y_k)} \lt 1##. So if we define ##A_k \equiv \frac {x_k} {(x_k+y_k)}##, we simply need to prove that ##\prod_{k=1}^n {A_k}^{1/n} + \prod_{k=1}^n {(1-A_k)}^{1/n} \le 1##

For positive integer ##n## and when ##A_1, A_2,\ldots,A_n## are all positive numbers in the range ##(0,1)##, we prove by induction that ##Y_n = \prod_{k=1}^n {A_k}^{1/n} + \prod_{k=1}^n {(1 - A_k)}^{1/n} \le 1##.
(Eq. 1)​
The base case is ##n=1## for which the equation reduces to ##A_1 + (1-A_1) = 1## and hence the condition is true. We assume that that the condition holds true for values of ##n## up to ##N## for some positive integer ##N##, i.e. ##Y_n \le 1 \ \forall n \in \{1, .., N-1\}## where ##N \ge 2##.

For ##n = N##, we have ##Y_n = Y_N = \prod_{k=1}^N {A_k}^{1/N} + \prod_{k=1}^N {(1 - A_k)}^{1/N} = \\
A_N \prod_{k=1}^{N-1} {A_k}^{1/N} + (1-A_N)\prod_{k=1}^{N-1} (1-A_k)^{1/N}##
(Eq. 2)​

For positive integer ##n## greater than 1, we prove the following identity: ##ax^{1/n} + b(1-x)^{1/n} \le (a^{n/(n-1)}+b^{n/(n-1)})^{(n-1)/n}## for any positive numbers ##a, b##, and ##0 \lt x \lt 1##.

Let ##y = ax^{1/n} + b(1-x)^{1/n}##. To find the maximum possible value of ##y## for a fixed ##a, b##, we find the value of ##x## for which ##y' = \frac {dy} {dx} = 0##. Solving this equation, we get ##\frac {a} {n} x^{1/n - 1} - \frac {b} {n} (1-x)^{1/n - 1} = 0 \Rightarrow \frac {1} {x} - 1 = (\frac {a} {b})^{\frac {n} {1-n}} \Rightarrow x = \frac {a^{\frac {n} {n-1}}} {a^{\frac {n} {n-1}} + b^{\frac {n} {n-1}}}##. Substituting this value of ##x## in the original equation for ##y##, we get ##y_{max} = (a^{\frac {n} {n-1}} + b^{\frac {n} {n-1}})^{\frac {n-1} {n}} \Rightarrow y \le (a^{n/(n-1)}+b^{n/(n-1)})^{(n-1)/n}##
(Eq. 3)​

Going back to (Eq. 2), we note that ##Y_N## is of the form ##ax^{1/n} + b(1-x)^{1/n}## (with ##A_N## corresponding to ##x##) for which we proved an identity (Eq. 3). Applying that identity, we get ##Y_N \le ((\prod_{k=1}^{N-1} {A_k}^{1/N})^{N / (N-1)} + (\prod_{k=1}^{N-1} {(1-A_k)}^{1/N})^{N / (N-1)})^{(N-1)/N} = \\
((\prod_{k=1}^{N-1} {A_k}^{1/(N-1)}) + (\prod_{k=1}^{N-1} {(1-A_k)}^{1/(N-1)}))^{(N-1)/N} \\
\Rightarrow Y_N \le {Y_{N-1}}^{(N-1)/N}##

Since by the inductive assumption ##Y_{N-1} \le 1##, we get ##{Y_{N-1}}^{(N-1)/N} \le 1 \Rightarrow Y_N \le 1##

Hence proved.

Mentor
I'm almost sure that it can be done this way, although my proof is significantly shorter by using AM ##\geq## GM.

15. Let ##\mathbf{x}=(x_1,\ldots ,x_n)\, , \,\mathbf{y}=(y_1,\ldots ,y_n)## be tuples of positive numbers. Prove

##\prod_{k=1}^n(x_k+y_k)^{1/n} ≥ \prod_{k=1}^n {x_k}^{1/n} + \prod_{k=1}^n {y_k}^{1/n}##​

To prove the given inequality, we prove its equivalent (obtained by the dividing both sides of the original inequality by the LHS), ##1 ≥ \prod_{k=1}^n {\frac {x_k} {(x_k+y_k)}}^{1/n} + \prod_{k=1}^n {\frac {y_k} {(x_k+y_k)}}^{1/n}##. Since ##x_k, y_k## are positive numbers, we have ##0 \lt \frac {x_k} {(x_k+y_k)} \lt 1## and ##0 \lt \frac {y_k} {(x_k+y_k)} \lt 1##. So if we define ##A_k \equiv \frac {x_k} {(x_k+y_k)}##, we simply need to prove that ##\prod_{k=1}^n {A_k}^{1/n} + \prod_{k=1}^n {(1-A_k)}^{1/n} \le 1##

For positive integer ##n## and when ##A_1, A_2,\ldots,A_n## are all positive numbers in the range ##(0,1)##, we prove by induction that ##Y_n = \prod_{k=1}^n {A_k}^{1/n} + \prod_{k=1}^n {(1 - A_k)}^{1/n} \le 1##.
(Eq. 1)​
The base case is ##n=1## for which the equation reduces to ##A_1 + (1-A_1) = 1## and hence the condition is true. We assume that that the condition holds true for values of ##n## up to ##N## for some positive integer ##N##, i.e. ##Y_n \le 1 \ \forall n \in \{1, .., N-1\}## where ##N \ge 2##.

For ##n = N##, we have ##Y_n = Y_N = \prod_{k=1}^N {A_k}^{1/N} + \prod_{k=1}^N {(1 - A_k)}^{1/N} = \\
A_N \prod_{k=1}^{N-1} {A_k}^{1/N} + (1-A_N)\prod_{k=1}^{N-1} (1-A_k)^{1/N}##
(Eq. 2)​
No, it is ##Y_N=
A_N^{1/N} \prod_{k=1}^{N-1} {A_k}^{1/N} + (1-A_N)^{1/N}\prod_{k=1}^{N-1} (1-A_k)^{1/N}##
For positive integer ##n## greater than 1, we prove the following identity: ##ax^{1/n} + b(1-x)^{1/n} \le (a^{n/(n-1)}+b^{n/(n-1)})^{(n-1)/n}## for any positive numbers ##a, b##, and ##0 \lt x \lt 1##.

Let ##y = ax^{1/n} + b(1-x)^{1/n}##. To find the maximum possible value of ##y## for a fixed ##a, b##, we find the value of ##x## for which ##y' = \frac {dy} {dx} = 0##. Solving this equation, we get ##\frac {a} {n} x^{1/n - 1} - \frac {b} {n} (1-x)^{1/n - 1} = 0 \Rightarrow \frac {1} {x} - 1 = (\frac {a} {b})^{\frac {n} {1-n}} \Rightarrow x = \frac {a^{\frac {n} {n-1}}} {a^{\frac {n} {n-1}} + b^{\frac {n} {n-1}}}##.

I get ##x=\left(1+\left(\dfrac{a}{b}\right)^{\frac{n}{1-n}}\right)^{-1}=\left(\dfrac{b^{\frac{n}{1-n}}+a^{\frac{n}{1-n}}}{b^{\frac{n}{1-n}}}\right)^{-1}=\dfrac{b^{\frac{n}{1-n}}}{a^{\frac{n}{1-n}}+b^{\frac{n}{1-n}}}## and now you lost me ... with a wrong numerator, a sign error in the exponent, the missing brackets in ##(1/n)-1## and the missing power of ##A_N## in mind, looking for a reason why it is a maximum and not a minimum ...

Substituting this value of ##x## in the original equation for ##y##, we get ##y_{max} = (a^{\frac {n} {n-1}} + b^{\frac {n} {n-1}})^{\frac {n-1} {n}} \Rightarrow y \le (a^{n/(n-1)}+b^{n/(n-1)})^{(n-1)/n}##
(Eq. 3)​

Going back to (Eq. 2), we note that ##Y_N## is of the form ##ax^{1/n} + b(1-x)^{1/n}## (with ##A_N## corresponding to ##x##) for which we proved an identity (Eq. 3). Applying that identity, we get ##Y_N \le ((\prod_{k=1}^{N-1} {A_k}^{1/N})^{N / (N-1)} + (\prod_{k=1}^{N-1} {(1-A_k)}^{1/N})^{N / (N-1)})^{(N-1)/N} = \\
((\prod_{k=1}^{N-1} {A_k}^{1/(N-1)}) + (\prod_{k=1}^{N-1} {(1-A_k)}^{1/(N-1)}))^{(N-1)/N} \\
\Rightarrow Y_N \le {Y_{N-1}}^{(N-1)/N}##

Since by the inductive assumption ##Y_{N-1} \le 1##, we get ##{Y_{N-1}}^{(N-1)/N} \le 1 \Rightarrow Y_N \le 1##

Hence proved.

I'm almost sure that it can be done this way, although my proof is significantly shorter by using AM ##\geq## GM.

No, it is ##Y_N=
A_N^{1/N} \prod_{k=1}^{N-1} {A_k}^{1/N} + (1-A_N)^{1/N}\prod_{k=1}^{N-1} (1-A_k)^{1/N}##

I get ##x=\left(1+\left(\dfrac{a}{b}\right)^{\frac{n}{1-n}}\right)^{-1}=\left(\dfrac{b^{\frac{n}{1-n}}+a^{\frac{n}{1-n}}}{b^{\frac{n}{1-n}}}\right)^{-1}=\dfrac{b^{\frac{n}{1-n}}}{a^{\frac{n}{1-n}}+b^{\frac{n}{1-n}}}## and now you lost me ... with a wrong numerator, a sign error in the exponent, the missing brackets in ##(1/n)-1## and the missing power of ##A_N## in mind, looking for a reason why it is a maximum and not a minimum ...

Sorry, I missed adding the exponent (##1/N##) to ##A_N## and ##(1-A_N)## in that expression for ##Y_N##... lost while translating the derivation I had worked on paper to LaTeX needed for the post. The value of ##x## that you have obtained is in fact equal to what I had stated earlier. Since ##\alpha^{\frac {p} {-q}} \equiv \dfrac {1} {\alpha^{\frac {p} {q}}}##, the expression that you obtained, ##\dfrac{b^{\frac{n}{1-n}}}{a^{\frac{n}{1-n}}+b^{\frac{n}{1-n}}}##, can be rewritten as ##\dfrac {\frac {1} {b^{n/(n-1)}}} {\frac {1} {a^{n/(n-1)}} + \frac {1} {b^{n/(n-1)}}}## which in turn can be simplified to ##\dfrac {{\frac {1} {b^{n/(n-1)}}} a^{n/(n-1)} b^{n/(n-1)}} {b^{n/(n-1)} + a^{n/(n-1)}} = \dfrac {a^{n/(n-1)}} {a^{n/(n-1)} + b^{n/(n-1)}}##, the same as what I had derived in the previous post.

To prove that the above value of ##x## corresponds to the maximum value of ##y \equiv ax^{1/n} + b(1-x)^{1/n}##, we take the second derivative of ##y## and show that it is negative throughout the range of valid values of ##x##, i.e. for any ##x \in (0, 1)##.

##y'' = \dfrac{a}{n} \left(\dfrac{1}{n} - 1 \right)x^{(\frac{1}{n} - 2)} + \dfrac{b}{n} \left(\dfrac{1}{n} - 1 \right)(1-x)^{(\frac{1}{n} - 2)} \\
= \left( \frac {1-n} {n^2}\right) \left(ax^{(\frac{1}{n} - 2)} + b(1-x)^{(\frac{1}{n} - 2)} \right)##.

Since as per the original preconditions of the identity ##a, b## are positive numbers, ##0 < x < 1## (which also implies ##0 < (1-x) < 1##, and ##n \gt 1##, we see that ##\left( \frac {1-n} {n^2}\right)## in the above expression for ##y''## must be negative whereas ##\left(ax^{(\frac{1}{n} - 2)} + b(1-x)^{(\frac{1}{n} - 2)} \right)## must be positive implying that the product and hence ##y''## is negative for all valid values of ##x##. Hence the value of ##x## in that valid range where ##y'## becomes 0 must correspond to a maximum (for ##y##) and not a minimum (since a negative value of ##y''## at some ##x = x_0## means that the slope of ##y## w.r.t. ##x## must be positive for values of ##x## that are in the left-side vicinity of ##x_0## and the slope must be negative for values of ##x > x_0## in the right-side vicinity of ##x_0##)

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Mentor
You left a bit too many steps to the reader, @Not anonymous. You can't tell me you made all of them in mind, so how can you expect we do? Let me see.

We have
$$Y_N=A_N^{1/N} \cdot \prod_{k=1}^{N-1}A_k^{1/N} + (1-A_N)^{1/N} \cdot \prod_{k=1}^{N-1}(1-A_k)^{1/N}$$
and
$$y= ax^{1/n}+b(1-x)^{1/n} \leq \left(a^{\dfrac{n}{n-1}}+b^{\dfrac{n}{n-1}}\right)^{\dfrac{n-1}{n}}$$
which is with ##x=A_N##
\begin{align*}
Y_N & \leq \left[ \left(\prod_{k=1}^{N-1}A_k\right)^{\dfrac{1}{N-1}} + \left(\prod_{k=1}^{N-1}(1-A_k)\right)^{\dfrac{1}{N-1}} \right]^{\dfrac{N-1}{N}}\\
&= Y_{N-1}^{\dfrac{N-1}{N}} \stackrel{\text{ I.H. }}{\leq} 1^{\dfrac{N-1}{N}}= 1
\end{align*}

O.k. understood now. Sorry, I had to write quite a bit and found it easier to do it here than on paper with all the quotient exponents.

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Not anonymous