Math Challenge - December 2019

[Fred Wright]

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.

 

[fresh_42]

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.

 

[Not anonymous]

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.

Spoiler: Question 14

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 $X_{min}(1/3) = 3/2 \Rightarrow X \geq 3/2$.

 

[fresh_42]

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

Spoiler: Question 14

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 high school?

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.

 

[Not anonymous]

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

 

[Not anonymous]

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

Spoiler

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.

 

[fresh_42]

Spoiler

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 $B$.

 

[Not anonymous]

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.

Spoiler: Question 12

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.

 

[fresh_42]

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!

 

[Not anonymous]

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

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

Spoiler: Question 15

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.

 

[fresh_42]

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.

 

[Not anonymous]

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 $x_0$)

 

[fresh_42]

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.