Challenge Intermediate Math Challenge - May 2018

[nuuskur]

Complex analysis a.k.a black magic

 

[Greg Bernhardt]

Only 4 left to solve and we have 2.5 weeks left in May. Awesome job everyone!

 

[fresh_42]

First day open to all. Open:
1.) Determinants.
7.) Series.
9.) Jacobi Identity.
10.) Integral.

 

[julian]

Is the solution of problem 10 this:

Spoiler: SOLUTION

By the residue theorem:

$I = 2 \pi i \lim_{z \rightarrow 0} e^{kz} = 2 \pi i .$

Then by the change of variables

$z = e^{i \theta} \qquad dz = i e^{i \theta} d \theta$

The integral becomes

\begin{align}
I & = \int_{- \pi}^\pi e^{k e^{i \theta}} \frac{i e^{i \theta} d \theta}{e^{i \theta}}
\nonumber \\
& = i \int_{- \pi}^\pi e^{k \cos \theta + i k \sin \theta} d \theta
\nonumber \\
& = i \int_{- \pi}^\pi e^{k \cos \theta} e^{i k \sin \theta} d \theta
\nonumber \\
& = i \int_{- \pi}^\pi e^{k \cos \theta} [ \cos (k \sin \theta) + i \sin (k \sin \theta) ] d \theta
\nonumber
\end{align}

Using that $\sin (k \sin \theta)$ is an odd function in $\theta$, we have

$I = i \int_{- \pi}^\pi e^{k \cos \theta} \cos (k \sin \theta) d \theta .$

Comparing this to the first answer we got for $I$, we have

$\int_{- \pi}^\pi e^{k \cos \theta} \cos (k \sin \theta) d \theta = 2 \pi .$

 

[QuantumQuest]

Is the solution of problem 10 this

Yes @julian. Well done!

 

[julian]

I think I have solved problem 7.

Spoiler: SOLUTION

I used the generating function technique. Define

$f (x) = \sum_{n=1}^\infty \frac{(n!)^2}{n (2n)!} x^n = \sum_{n=1}^\infty a_n x^n$

The sum we are after will then be equal to $f (1)$.

Note that

$a_{n+1} = \frac{(n!)^2}{n (2n)!} \times \frac{n (n+1)^2}{(n+1) (2n+2) (2n+1)} = a_n \times \frac{n}{2 (2n +1)}$

or $2 (2n+1) a_{n+1} = n a_n$. In the following we use $a_1 = 1/2$. First

\begin{align}
x \frac{d}{dx} f (x) & = \sum_{n=1}^\infty n a_n x^n
\nonumber \\
& = \sum_{n=1}^\infty 2 (2n+1) a_{n+1} x^n
\nonumber
\end{align}

Then

\begin{align}
x^2 \frac{d}{dx} f (x) & = \sum_{n=1}^\infty 2 (2n+1) a_{n+1} x^{n+1}
\nonumber \\
& = 2 \sum_{m=2}^\infty (2m -1) a_m x^m
\nonumber \\
& = 2 \sum_{m=1}^\infty (2m -1) a_m x^m - 2 (2 \times 1 - 1) a_{1} x
\nonumber \\
& = 4 \sum_{m=1}^\infty m a_m x^m - 2 f (x) - x
\nonumber \\
& = 4 x \frac{d}{dx} f (x) - 2 f (x) - x
\nonumber
\end{align}

This gives the differential equation:

$(x^2 - 4x) \frac{d}{dx} f (x) + 2f (x) + x = 0$

or

$\frac{d}{dx} f (x) + \frac{2}{x^2 - 4x} f (x) = - \frac{1}{x - 4} .$

(with boundary condition $f (0) = 0$). Now this is the famililar form

$\frac{d}{dx} f (x) + p (x) f (x) = q (x)$

that can be solved using the integrating factor method which uses

\begin{align}
\frac{d}{dx} \Big[ e^{\nu (x)} f (x) \Big] & = e^{\nu (x)} \Big[ \frac{d}{dx} f (x) + p (x) f (x) \Big]
\nonumber \\
& = e^{\nu (x)} q (x)
\nonumber
\end{align}

where $\nu (x) = \int p (x) dx$. We will solve this with

$f (x) = e^{- \nu (x)} \int e^{\nu (x)} q(x) dx .$

In our case

$p (x) = - \frac{1}{2} \Big( \frac{1}{x} - \frac{1}{x-4} \Big) , \qquad q (x) = - \frac{1}{x-4}$

so that

\begin{align}
\nu (x) & = - \frac{1}{2} \int \Big( \frac{1}{x} - \frac{1}{x - 4} \Big) dx
\nonumber \\
& = - \frac{1}{2} ( \ln x - \ln (x-4))
\nonumber
\end{align}

meaning

$e^{+ \nu (x)} = \frac{(x-4)^{1/2}}{x^{1/2}} \qquad \mathrm{and} \qquad e^{- \nu (x)} = \frac{x^{1/2}}{(x-4)^{1/2}}$

and

\begin{align}
f (x) & = - \frac{x^{1/2}}{(x-4)^{1/2}} \int_0^x \frac{(x-4)^{1/2}}{x^{1/2}} \frac{1}{x-4} dx
\nonumber \\
& = i \frac{x^{1/2}}{(4-x)^{1/2}} \int_0^x \frac{dx}{(x^2 - 4x)^{1/2}} dx
\nonumber
\end{align}

(choosing the lower limit to be $0$ will ensure that $f (0) = 0$). We now take $x = 1$, then

\begin{align}
\sum_{n=1}^\infty \frac{(n!)^2}{n (2n)!} & = f (1)
\nonumber \\
& = i \frac{1}{\sqrt{3}} \int_0^1 \frac{dx}{(x^2 - 4 x)^{1/2}}
\nonumber
\end{align}

We evaluate the above integral. Write $x^2 - 4 x = [x - 2]^2 - 4$. Then with the substitution $u = x - 2$ we obtain

\begin{align}
\int_0^1 \frac{dx}{(x^2 - 4 x)^{1/2}} & = \int_{-2}^{-1} \frac{du}{(u^2 - 2^2)^{1/2}}
\nonumber \\
& = \Big[ \cosh^{-1} (u/2) \Big]_{-2}^{-1}
\nonumber \\
& = \cosh^{-1} (-1/2) - \cosh^{-1} (-1) .
\nonumber
\end{align}

Using the formula $\cosh^{-1} x = \ln [x + \sqrt{x+1} \sqrt{x-1}]$, we have

\begin{align}
\cosh^{-1} (-1/2) - \cosh^{-1} (-1) & = \ln [-1/2 + i \sqrt{3}/2] - \ln (-1)
\nonumber\\
& = i \frac{2}{3} \pi - i \pi
\nonumber\\
& = - i \frac{1}{3} \pi
\nonumber
\end{align}

Accordingly

\begin{align}
\sum_{n=1}^\infty \frac{(n!)^2}{n (2n)!} & = i \frac{1}{\sqrt{3}} \times -i \frac{\pi}{3}
\nonumber \\
& = \frac{\pi}{3 \sqrt{3}} .
\nonumber
\end{align}

 

[fresh_42]

I think I have solved problem 7.

Wow! It looks good at first glance, i.e. the result is correct and I couldn't find flaws. There is a simpler solution using a known series expansion, but this one is really good.

 

[julian]

Thanks fresh_42. A thing that needs to be checked is if we are allowed to differentiate the series term by term. This is easy to establish.

There is the theorem:

The derivative of the series sum $f (x) = \sum_i^\infty u_n (x)$ equals the sum of the individual term derivatives,

$\frac{d}{dx} f (x) = \sum_{n=1}^\infty \frac{d}{dx} u_n (x) ,$

provided the following conditions hold:

$u_n (x) \quad \mathrm{and} \quad \frac{d u_n (x)}{dx} \quad \mathrm{are \; continuous \; in} \; [a,b]$

$\sum_{n=1}^\infty \frac{d u_n (x)}{dx} \quad \mathrm{is \; uniformally \; convergent \; in} \; [a,b] .$

In our case it is obvious that $u_n (x) = a_n x^n$ and $\frac{u_n (x)}{dx} = n a_n x^{n-1}$ are continuous in $[0,1]$. So we just have to check that $\sum_{n=1}^\infty \frac{d u_n (x)}{dx}$ is uniformally convergent.

We can do this using the Weierstrass M test. This states that if we can construct a series of numbers $\sum_1^\infty M_n$, in which $M_n \geq |v_n (x)|$ for all $x \in [a,b]$ and $\sum_1^\infty M_n$ is convergent, the series $\sum_1^\infty v_n (x)$ will be uniformally convergent in $[a,b]$.

In our case an obvious candidate for the $M_n$'s are $n a_n = \frac{(n!)^2}{(2n)!}$. The ratio test tells you the series $\sum_1^\infty M_n$ is convergent:

$\lim_{n \rightarrow \infty} \frac{(n+1) a_{n+1}}{n a_n} = \lim_{n \rightarrow \infty} \frac{(n+1)^2}{(2n+2) (2n+1)} = \frac{1}{4} < 1 .$

We then just need to note that

$M_n = n a_n \geq |n a_n x^{n-1}| = \Big| \frac{d u_n (x)}{dx} \Big| \quad \mathrm{for \; all} \;\; x \in [0,1] .$Also, it might have been "nicer" to write

$f (x) = \frac{x^{1/2}}{(4-x)^{1/2}} \int_0^x \frac{dx}{(4 - x^2)^{1/2}}$

and

\begin{align}
f (1) & = \frac{1}{\sqrt{3}} \int_0^1 \frac{dx}{(4x - x^2)^{1/2}}
\nonumber \\
& = \frac{1}{\sqrt{3}} \int_0^1 \frac{dx}{[2^2 - (x - 2)^2]^{1/2}}
\nonumber \\
& = \frac{1}{\sqrt{3}} \int_{-2}^{-1}\frac{du}{[2^2 - u^2]^{1/2}}
\nonumber \\
& = \frac{1}{\sqrt{3}} \Big[ \sin^{-1} (u/2) \Big]_{-2}^{-1}
\nonumber \\
& = \frac{1}{\sqrt{3}} \Big( \sin^{-1} (1) - \sin^{-1} (1/2) \Big)
\nonumber \\
& = \frac{1}{\sqrt{3}} \Big( \frac{\pi}{2} - \frac{\pi}{6} \Big)
\nonumber \\
& = \frac{\pi}{3 \sqrt{3}}
\nonumber
\end{align}

but it is equivalent.

 

[julian]

I think I have solved problem 1:

I split the proof into the parts:

Part (a) A few facts about real skew symmetric matrices.
Part (b): Proof for $n$ even.
(i) Looking at case $n = 2$.
(ii) Proving a key inequality. This will prove case $n=2$
(iii) Proving case for general even $n$ (then easy).
Part (c) Case of odd $n$.
(i) Proving case for $n = 3$ (easy because of part (b)).
(iii) Proving case for general odd $n$ (easy because of part (b)).

Spoiler: part (a) and (b)(i)

Part (a):

Few facts about real skew matrices:

They are normal: $A A^\dagger = A A^T = -A A = A^T A = A^\dagger A$ and so the spectral theorem holds. There is a unitary matrix $U$ such that $U^\dagger A U = D$ where $D$ is a diagonal matrix. The entries of the diagonal of $D$ are the eigenvalues of $A$.
The eigenvalues are pure imaginary.
As the coefficients of the characteristic polynomial, $\det (A - \lambda I) = 0$, are real the eigenvalues come in conjugate pairs. If the dimension, $n$, of the matrix $A$ is odd then 0 must be one of the eigenvalues.

Part (b) (i):

We first take the simplest case of even $n$: $n = 2$. We can write

$U^\dagger A U = D = \begin{pmatrix} \lambda & 0 \\ 0 & - \lambda \end{pmatrix} .$

First consider

\begin{align}
& \prod_{i=1}^k \det \big( A + x_i I \big) =
\nonumber \\
& = \det \big[ \big( A + x_1 I \big) \big( A + x_2 I \big) \dots \big( A + x_k I \big) \big]
\nonumber \\
& = \det \big[ U^{-1} \big( A + x_1 I \big) U U^{-1} \big( A + x_2 I \big) U U^{-1} \dots U U^{-1} \big( A + x_k
I \big) U \big]
\nonumber \\
& = \det \Big[
\begin{pmatrix}
\lambda + x_1 & 0 \\
0 & - \lambda + x_1
\end{pmatrix}
\begin{pmatrix}
\lambda + x_2 & 0 \\
0 & - \lambda + x_2
\end{pmatrix}
\dots
\begin{pmatrix}
\lambda + x_k & 0 \\
0 & - \lambda + x_k
\end{pmatrix}
\Big]
\nonumber \\
& = \det
\nonumber \\
&
\begin{pmatrix}
(\lambda + x_1) (\lambda + x_2) \dots (\lambda + x_k) & 0 \\
0 & (- \lambda + x_1) (- \lambda + x_2) \dots (- \lambda + x_k)
\end{pmatrix}
\nonumber \\
& = (- \lambda^2 + x_1^2) (- \lambda^2 + x_2^2) \dots (- \lambda^2 + x_k^2)
\nonumber \\
& = (\theta^2 + x_1^2) (\theta^2 + x_2^2) \dots (\theta^2 + x_k^2) \qquad \qquad \qquad \qquad \qquad \qquad (1)
\nonumber
\end{align}

where we have introduced $\lambda = i \theta$ where $\theta$ is real.

Now consider

\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) =
\nonumber \\
& =
\det \Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}^k
\Big]
\nonumber \\
& = \det \Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}
\Big]^k
\nonumber \\
& = \big[ - \lambda^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber \\
&= \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\qquad \qquad (2)
\nonumber
\end{align}

Comparing eq (1) and eq (2), we see that proving the main result for $n = 2$ then amounts to proving the inequality:

$(\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k \qquad \qquad \qquad \qquad (3)$

We prove this, for general value of $\theta$, in the next subsection. See next spoiler!

Spoiler: Part (b) (ii)

Part (b) (ii):

We wish to prove (3):

$(\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k$

The proof is an essentially inductive argument. Our base case $k = 2$ is easy:

\begin{align}
(\theta^2 + x_1^2) (\theta^2 + x_2^2) & = \theta^4 + \theta^2 x_1^2 + \theta^2 x_2^2 + x_1^2 x_2^2
\nonumber \\
& \geq \theta^4 + 2 \theta^2 (x_1 x_2) + x_1^2 x_2^2
\nonumber \\
& = [\theta^2 + (x_1^2 x_2^2)^{1/2}]^2
\nonumber
\end{align}

where we used $a^2 + b^2 \geq 2 ab$.

Next we prove that whenever the result holds for $k$, it holds for $2k$ as well. That is, we'll first prove the result for powers of $2: k = 2, 4, 8, 16, \dots$. Assume we know the result holds for some $k$. Now consider $2k$ positive numbers $x_1^2, \dots , x_k^2$ and $y_1^k , \dots y_k^2$. We use the induction hypothesis and the base case to find

\begin{align}
& (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) (\theta^2 + y_1^2) \dots (\theta^2 + y_k^2)
\nonumber \\
& \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k [\theta^2 + (y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& = [\theta^4 + \theta^2 (x_1^2 \dots x_k^2)^{1/k} + \theta^2 (y_1^2 \dots y_k^2)^{1/k} + (x_1^2 \dots x_k^2
y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& \geq [\theta^4 + 2 \theta^2 (x_1^2 \dots x_k^2 y_1^2 \dots y_k^2)^{1/2k} + (x_1^2 \dots x_k^2
y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& = [\theta^2 + (x_1^2 \dots x_k^2 y_1^2 \dots y_k^2)^{1/2k}]^{2k}
\nonumber
\end{align}

where we used $a^2 + b^2 \geq 2 ab$. This is the required result. We now know the theorem to be true for infinitely many $k$.

Next we prove that whenever the result is true for $k$, it's also true for $k - 1$. This will prove the result for all the in-between integers. Let $k \geq 4$ and assume the result holds for $k$. Consider the $k-1$ positive numbers $x_1^2 , x_2^2 , \dots , x_{k-1}^2$. Define $x_k^2$ to be $(x_1^2 x_2^2 \dots x_{k-1}^2)^{1/(k-1)}$. We then have

\begin{align}
& (\theta^2 + x_1^2) (\theta^2 + x_2^2) \dots (\theta^2 + x_{k-1}^2) (\theta^2 + x_k^2)
\nonumber \\
& \geq [\theta^2 + (x_1^2 \dots x_{k-1}^2 x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + \{ (x_1^2 \dots x_{k-1}^2)^{1/(k-1)} \}^{(k-1)/k} \; (x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + \big( x_k^2 \big)^{(k-1)/k} \; (x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + x_k^2]^k
\nonumber
\end{align}

Rearranging we have

\begin{align}
& (\theta^2 + x_1^2) (\theta^2 + x_2^2) \dots (\theta^2 + x_{k-1}^2) \geq [\theta^2 + x_k^2]^{k-1}
\nonumber \\
& \qquad \equiv [\theta^2 + (x_1^2 x_2^2 \dots x_{k-1}^2)^{1 / (k-1)} ]^{k-1}
\nonumber
\end{align}

which is the required result.

Which means we have established (3) and proven the main result for $n = 2$, i.e.,

$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n = 2 .$

We will prove the main result for arbitrary even $n$ in the next subsection. See next spoiler!

Spoiler: Part (b) (iii)

Part (b) (iii):

We now prove the main result for general even $n$. We can write

$U^\dagger A U = D = \begin{pmatrix} \lambda_1 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & - \lambda_1 & 0 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & \lambda_2 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & - \lambda_2& \dots & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \dots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \dots & \dots & \lambda_{n/2} & 0 \\ 0 & 0 & 0 & 0 & \dots & \dots & 0 & - \lambda_{n/2} \\ \end{pmatrix} .$

So that

\begin{align}
& \prod_{i=1}^k \det (A - x_i I)
\nonumber \\
& = \prod_{i=1}^k \det
\nonumber \\
&
\begin{pmatrix}
\lambda_1 + x_i & 0 & \dots & \dots & 0 & 0 \\
0 & - \lambda_1 + x_i & \dots & \dots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
\vdots & \vdots & \dots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & \dots & \lambda_{n/2} + x_i & 0 \\
0 & 0 & \dots & \dots & 0 & - \lambda_{n/2} + x_i \\
\end{pmatrix}
\nonumber \\
& =
\Big( \prod_{i=1}^k
\det
\begin{pmatrix}
\lambda_1 + x_i & 0 \\
0 & - \lambda_1 + x_i
\end{pmatrix} \Big)
\dots
\Big( \prod_{i=1}^k
\begin{pmatrix}
\lambda_{n/2} + x_i & 0 \\
0 & - \lambda_{n/2} + x_i
\end{pmatrix} \Big)
\nonumber \\
& = \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big) \dots \big( \prod_{i=1}^k(\theta_{n/2}^2 + x_i^2) \big)
\nonumber
\end{align}

where we have introduced $\lambda_l = i \theta_l$ where $\theta_l$ is real..

Next consider

$\qquad \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k$
$\quad = \det \begin{pmatrix} \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & 0 & \dots & \dots \\ 0 & - \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & \dots & \dots \\ \vdots & \vdots & \ddots & \vdots \\ \vdots & \vdots & \dots & \ddots \end{pmatrix}^k$
$\quad = \prod_{l=1}^{n/2} \det \Big[ \begin{pmatrix} \lambda_l + (x_1 x_2 \dots x_k)^{1/k} & 0 \\ 0 & - \lambda_l + (x_1 x_2 \dots x_k)^{1/k} \end{pmatrix}^k \Big]$
$\quad = \big[ \theta_1^2 + ( x_1^2 x_2^2 \dots x_k^2 )^{1/k} \big]^k \dots \big[ \theta_{n/2}^2 + ( x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k$

We easily have from (3) that

\begin{align}
& \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{n/2}^2 + x_i^2) \big)
\geq
\nonumber \\
& \qquad \qquad \qquad \qquad \qquad
\big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{\frac{1}{k}} \big]^k
\dots
\big[ \theta_{n/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{\frac{1}{k}} \big]^k
\nonumber
\end{align}

which proves the main result for even $n$:

$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n \; \mathrm{even} .$

In the next section we prove the main result for odd $n$. See next spoiler!

Spoiler: part (c)

Part (c) (i):

We now consider the case $n = 3$. We write

$U^\dagger A U = D = \begin{pmatrix} 0 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & - \lambda \end{pmatrix} .$

First consider

$\prod_{i=1}^k \det \big( A + x_i I \big) =$
$= \det \big[ U^{-1} \big( A + x_1 I \big) U U^{-1} \dots U U^{-1} \big( A + x_k I \big) U \big]$
$\quad = \det \Big[ \begin{pmatrix} x_1 & 0 & 0 \\ 0 & \lambda + x_1 & 0 \\ 0 & 0 & - \lambda + x_1 \end{pmatrix} \dots \begin{pmatrix} x_k & 0 & 0 \\ 0 & \lambda + x_k & 0 \\ 0 & 0 & - \lambda + x_k \end{pmatrix} \Big]$
$= \det$
$\begin{pmatrix} x_1 x_2 \dots x_k & 0 & 0 \\ 0 & (\lambda + x_1) \dots (\lambda + x_k) & 0 \\ 0 & 0 & (- \lambda + x_1) \dots (- \lambda + x_k) \end{pmatrix}$
$= (x_1 x_2 \dots x_k) \times$
$\quad \det \begin{pmatrix} (\lambda + x_1) \dots (\lambda + x_k) & 0 \\ 0 & (- \lambda + x_1) \dots (- \lambda + x_k) \end{pmatrix} .$
$= (x_1 x_2 \dots x_k) (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2)$

Now consider

\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k =
\nonumber \\
& = \det
\Big[
\begin{pmatrix}
(x_1 x_2 \dots x_k)^{\frac{1}{k}} & 0 & 0 \\
0 & \lambda + (x_1 x_2 \dots x_k)^{\frac{1}{k}} & 0 \\
0 & 0 & - \lambda + (x_1 x_2 \dots x_k)^{\frac{1}{k}}
\end{pmatrix}
\Big]^k
\nonumber \\
& = (x_1 x_2 \dots x_k) \det
\Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}
\Big]^k
\nonumber \\
& = (x_1 x_2 \dots x_k ) \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}

As the OP's question assumes that $\{ x_1, \dots , x_k \}$ are positive numbers, and using the result (3), we

have:

$(x_1 x_2 \dots x_k) (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq (x_1 x_2 \dots x_k ) \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k$

which establishes the main result for $n = 3$:

$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n = 3 .$

Part (c) (ii):

We now turn to the general case of any odd $n$. We can write

$U^\dagger A U = D = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & 0 & - \lambda_1 & 0 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & \lambda_2 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & 0 & - \lambda_2& \dots & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \dots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \dots & \dots & \lambda_n & 0 \\ 0 & 0 & 0 & 0 & 0 & \dots & \dots & 0 & - \lambda_n \\ \end{pmatrix} .$

First consider

\begin{align}
& \prod_{i=1}^k \det \big( A + x_i I \big) =
\nonumber \\
& = \prod_{i=1}^k \det
\nonumber \\
&
\begin{pmatrix}
x_i & 0 & 0 & \dots & 0 & 0 \\
0 & \lambda_1 + x_i & 0 & \dots & 0 & 0 \\
0 & 0 & - \lambda_1 +x_i & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \dots & \lambda_{(n-1)/2}+ x_i & 0 \\
0 & 0 & 0 & \dots & 0 & - \lambda_{(n-1)/2} + x_i \\
\end{pmatrix}
\nonumber \\
& = (x_1 x_2 \dots x_k ) \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{(n-1)/2}^2 + x_i^2) \big)
\nonumber
\end{align}

Now consider

\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k =
\nonumber \\
& = \det
\begin{pmatrix}
(x_1 \dots x_k)^{\frac{1}{k}} & 0 & 0 & \dots \\
0 & \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & 0 & \dots \\
0 & 0 & - \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & \dots \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\end{pmatrix}^k
\nonumber \\
& =
(x_1 x_2 \dots x_k ) \big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\dots
\big[ \theta_{(n-1)/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}

We easily have from (3) that

\begin{align}
& (x_1 x_2 \dots x_k ) \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{(n-1)/2}^2 + x_i^2) \big)
\geq
\nonumber \\
&
(x_1 x_2 \dots x_k ) \big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\dots
\big[ \theta_{(n-1)/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}

which establishes the main result for general odd values of $n$:

$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad \mathrm {for \; odd \;} n .$Completeting the proof of problem 1.

 

[StoneTemplePython]

I think I have solved problem 1:

I split the proof into the parts:

Part (a) A few facts about real skew symmetric matrices.
Part (b): Proof for $n$ even.
(i) Looking at case $n = 2$.
(ii) Proving a key inequality. This will prove case $n=2$
(iii) Proving case for general even $n$ (then easy).
Part (c) Case of odd $n$.
(i) Proving case for $n = 3$ (easy because of part (b)).
(iii) Proving case for general odd $n$ (easy because of part (b)).

Spoiler: part (a) and (b)(i)

Part (a):

Few facts about real skew matrices:

They are normal: $A A^\dagger = A A^T = -A A = A^T A = A^\dagger A$ and so the spectral theorem holds. There is a unitary matrix $U$ such that $U^\dagger A U = D$ where $D$ is a diagonal matrix. The entries of the diagonal of $D$ are the eigenvalues of $A$.
The eigenvalues are pure imaginary.
As the coefficients of the characteristic polynomial, $\det (A - \lambda I) = 0$, are real the eigenvalues come in conjugate pairs. If the dimension, $n$, of the matrix $A$ is odd then 0 must be one of the eigenvalues.

Part (b) (i):

We first take the simplest case of even $n$: $n = 2$. We can write

$U^\dagger A U = D = \begin{pmatrix} \lambda & 0 \\ 0 & - \lambda \end{pmatrix} .$

First consider

\begin{align}
& \prod_{i=1}^k \det \big( A + x_i I \big) =
\nonumber \\
& = \det \big[ \big( A + x_1 I \big) \big( A + x_2 I \big) \dots \big( A + x_k I \big) \big]
\nonumber \\
& = \det \big[ U^{-1} \big( A + x_1 I \big) U U^{-1} \big( A + x_2 I \big) U U^{-1} \dots U U^{-1} \big( A + x_k
I \big) U \big]
\nonumber \\
& = \det \Big[
\begin{pmatrix}
\lambda + x_1 & 0 \\
0 & - \lambda + x_1
\end{pmatrix}
\begin{pmatrix}
\lambda + x_2 & 0 \\
0 & - \lambda + x_2
\end{pmatrix}
\dots
\begin{pmatrix}
\lambda + x_k & 0 \\
0 & - \lambda + x_k
\end{pmatrix}
\Big]
\nonumber \\
& = \det
\nonumber \\
&
\begin{pmatrix}
(\lambda + x_1) (\lambda + x_2) \dots (\lambda + x_k) & 0 \\
0 & (- \lambda + x_1) (- \lambda + x_2) \dots (- \lambda + x_k)
\end{pmatrix}
\nonumber \\
& = (- \lambda^2 + x_1^2) (- \lambda^2 + x_2^2) \dots (- \lambda^2 + x_k^2)
\nonumber \\
& = (\theta^2 + x_1^2) (\theta^2 + x_2^2) \dots (\theta^2 + x_k^2) \qquad \qquad \qquad \qquad \qquad \qquad (1)
\nonumber
\end{align}

where we have introduced $\lambda = i \theta$ where $\theta$ is real.

Now consider

\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) =
\nonumber \\
& =
\det \Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}^k
\Big]
\nonumber \\
& = \det \Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}
\Big]^k
\nonumber \\
& = \big[ - \lambda^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber \\
&= \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\qquad \qquad (2)
\nonumber
\end{align}

Comparing eq (1) and eq (2), we see that proving the main result for $n = 2$ then amounts to proving the inequality:

$(\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k \qquad \qquad \qquad \qquad (3)$

We prove this, for general value of $\theta$, in the next subsection. See next spoiler!

Spoiler: Part (b) (ii)

Part (b) (ii):

We wish to prove (3):

$(\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k$

The proof is an essentially inductive argument. Our base case $k = 2$ is easy:

\begin{align}
(\theta^2 + x_1^2) (\theta^2 + x_2^2) & = \theta^4 + \theta^2 x_1^2 + \theta^2 x_2^2 + x_1^2 x_2^2
\nonumber \\
& \geq \theta^4 + 2 \theta^2 (x_1 x_2) + x_1^2 x_2^2
\nonumber \\
& = [\theta^2 + (x_1^2 x_2^2)^{1/2}]^2
\nonumber
\end{align}

where we used $a^2 + b^2 \geq 2 ab$.

Next we prove that whenever the result holds for $k$, it holds for $2k$ as well. That is, we'll first prove the result for powers of $2: k = 2, 4, 8, 16, \dots$. Assume we know the result holds for some $k$. Now consider $2k$ positive numbers $x_1^2, \dots , x_k^2$ and $y_1^k , \dots y_k^2$. We use the induction hypothesis and the base case to find

\begin{align}
& (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) (\theta^2 + y_1^2) \dots (\theta^2 + y_k^2)
\nonumber \\
& \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k [\theta^2 + (y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& = [\theta^4 + \theta^2 (x_1^2 \dots x_k^2)^{1/k} + \theta^2 (y_1^2 \dots y_k^2)^{1/k} + (x_1^2 \dots x_k^2
y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& \geq [\theta^4 + 2 \theta^2 (x_1^2 \dots x_k^2 y_1^2 \dots y_k^2)^{1/2k} + (x_1^2 \dots x_k^2
y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& = [\theta^2 + (x_1^2 \dots x_k^2 y_1^2 \dots y_k^2)^{1/2k}]^{2k}
\nonumber
\end{align}

where we used $a^2 + b^2 \geq 2 ab$. This is the required result. We now know the theorem to be true for infinitely many $k$.

Next we prove that whenever the result is true for $k$, it's also true for $k - 1$. This will prove the result for all the in-between integers. Let $k \geq 4$ and assume the result holds for $k$. Consider the $k-1$ positive numbers $x_1^2 , x_2^2 , \dots , x_{k-1}^2$. Define $x_k^2$ to be $(x_1^2 x_2^2 \dots x_{k-1}^2)^{1/(k-1)}$. We then have

\begin{align}
& (\theta^2 + x_1^2) (\theta^2 + x_2^2) \dots (\theta^2 + x_{k-1}^2) (\theta^2 + x_k^2)
\nonumber \\
& \geq [\theta^2 + (x_1^2 \dots x_{k-1}^2 x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + \{ (x_1^2 \dots x_{k-1}^2)^{1/(k-1)} \}^{(k-1)/k} \; (x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + \big( x_k^2 \big)^{(k-1)/k} \; (x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + x_k^2]^k
\nonumber
\end{align}

Rearranging we have

\begin{align}
& (\theta^2 + x_1^2) (\theta^2 + x_2^2) \dots (\theta^2 + x_{k-1}^2) \geq [\theta^2 + x_k^2]^{k-1}
\nonumber \\
& \qquad \equiv [\theta^2 + (x_1^2 x_2^2 \dots x_{k-1}^2)^{1 / (k-1)} ]^{k-1}
\nonumber
\end{align}

which is the required result.

Which means we have established (3) and proven the main result for $n = 2$, i.e.,

$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n = 2 .$

We will prove the main result for arbitrary even $n$ in the next subsection. See next spoiler!

Spoiler: Part (b) (iii)

Part (b) (iii):

We now prove the main result for general even $n$. We can write

$U^\dagger A U = D = \begin{pmatrix} \lambda_1 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & - \lambda_1 & 0 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & \lambda_2 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & - \lambda_2& \dots & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \dots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \dots & \dots & \lambda_{n/2} & 0 \\ 0 & 0 & 0 & 0 & \dots & \dots & 0 & - \lambda_{n/2} \\ \end{pmatrix} .$

So that

\begin{align}
& \prod_{i=1}^k \det (A - x_i I)
\nonumber \\
& = \prod_{i=1}^k \det
\nonumber \\
&
\begin{pmatrix}
\lambda_1 + x_i & 0 & \dots & \dots & 0 & 0 \\
0 & - \lambda_1 + x_i & \dots & \dots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
\vdots & \vdots & \dots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & \dots & \lambda_{n/2} + x_i & 0 \\
0 & 0 & \dots & \dots & 0 & - \lambda_{n/2} + x_i \\
\end{pmatrix}
\nonumber \\
& =
\Big( \prod_{i=1}^k
\det
\begin{pmatrix}
\lambda_1 + x_i & 0 \\
0 & - \lambda_1 + x_i
\end{pmatrix} \Big)
\dots
\Big( \prod_{i=1}^k
\begin{pmatrix}
\lambda_{n/2} + x_i & 0 \\
0 & - \lambda_{n/2} + x_i
\end{pmatrix} \Big)
\nonumber \\
& = \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big) \dots \big( \prod_{i=1}^k(\theta_{n/2}^2 + x_i^2) \big)
\nonumber
\end{align}

where we have introduced $\lambda_l = i \theta_l$ where $\theta_l$ is real..

Next consider

$\qquad \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k$
$\quad = \det \begin{pmatrix} \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & 0 & \dots & \dots \\ 0 & - \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & \dots & \dots \\ \vdots & \vdots & \ddots & \vdots \\ \vdots & \vdots & \dots & \ddots \end{pmatrix}^k$
$\quad = \prod_{l=1}^{n/2} \det \Big[ \begin{pmatrix} \lambda_l + (x_1 x_2 \dots x_k)^{1/k} & 0 \\ 0 & - \lambda_l + (x_1 x_2 \dots x_k)^{1/k} \end{pmatrix}^k \Big]$
$\quad = \big[ \theta_1^2 + ( x_1^2 x_2^2 \dots x_k^2 )^{1/k} \big]^k \dots \big[ \theta_{n/2}^2 + ( x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k$

We easily have from (3) that

\begin{align}
& \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{n/2}^2 + x_i^2) \big)
\geq
\nonumber \\
& \qquad \qquad \qquad \qquad \qquad
\big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{\frac{1}{k}} \big]^k
\dots
\big[ \theta_{n/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{\frac{1}{k}} \big]^k
\nonumber
\end{align}

which proves the main result for even $n$:

$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n \; \mathrm{even} .$

In the next section we prove the main result for odd $n$. See next spoiler!

Spoiler: part (c)

Part (c) (i):

We now consider the case $n = 3$. We write

$U^\dagger A U = D = \begin{pmatrix} 0 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & - \lambda \end{pmatrix} .$

First consider

$\prod_{i=1}^k \det \big( A + x_i I \big) =$
$= \det \big[ U^{-1} \big( A + x_1 I \big) U U^{-1} \dots U U^{-1} \big( A + x_k I \big) U \big]$
$\quad = \det \Big[ \begin{pmatrix} x_1 & 0 & 0 \\ 0 & \lambda + x_1 & 0 \\ 0 & 0 & - \lambda + x_1 \end{pmatrix} \dots \begin{pmatrix} x_k & 0 & 0 \\ 0 & \lambda + x_k & 0 \\ 0 & 0 & - \lambda + x_k \end{pmatrix} \Big]$
$= \det$
$\begin{pmatrix} x_1 x_2 \dots x_k & 0 & 0 \\ 0 & (\lambda + x_1) \dots (\lambda + x_k) & 0 \\ 0 & 0 & (- \lambda + x_1) \dots (- \lambda + x_k) \end{pmatrix}$
$= (x_1 x_2 \dots x_k) \times$
$\quad \det \begin{pmatrix} (\lambda + x_1) \dots (\lambda + x_k) & 0 \\ 0 & (- \lambda + x_1) \dots (- \lambda + x_k) \end{pmatrix} .$
$= (x_1 x_2 \dots x_k) (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2)$

Now consider

\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k =
\nonumber \\
& = \det
\Big[
\begin{pmatrix}
(x_1 x_2 \dots x_k)^{\frac{1}{k}} & 0 & 0 \\
0 & \lambda + (x_1 x_2 \dots x_k)^{\frac{1}{k}} & 0 \\
0 & 0 & - \lambda + (x_1 x_2 \dots x_k)^{\frac{1}{k}}
\end{pmatrix}
\Big]^k
\nonumber \\
& = (x_1 x_2 \dots x_k) \det
\Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}
\Big]^k
\nonumber \\
& = (x_1 x_2 \dots x_k ) \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}

As the OP's question assumes that $\{ x_1, \dots , x_k \}$ are positive numbers, and using the result (3), we

have:

$(x_1 x_2 \dots x_k) (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq (x_1 x_2 \dots x_k ) \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k$

which establishes the main result for general odd values of $n = 3$:

$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n = 3 .$

Part (c) (ii):

We now turn to the general case of any odd $n$. We can write

$U^\dagger A U = D = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & 0 & - \lambda_1 & 0 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & \lambda_2 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & 0 & - \lambda_2& \dots & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \dots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \dots & \dots & \lambda_n & 0 \\ 0 & 0 & 0 & 0 & 0 & \dots & \dots & 0 & - \lambda_n \\ \end{pmatrix} .$

First consider

\begin{align}
& \prod_{i=1}^k \det \big( A + x_i I \big) =
\nonumber \\
& = \prod_{i=1}^k \det
\nonumber \\
&
\begin{pmatrix}
x_i & 0 & 0 & \dots & 0 & 0 \\
0 & \lambda_1 + x_i & 0 & \dots & 0 & 0 \\
0 & 0 & - \lambda_1 +x_i & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \dots & \lambda_{(n-1)/2}+ x_i & 0 \\
0 & 0 & 0 & \dots & 0 & - \lambda_{(n-1)/2} + x_i \\
\end{pmatrix}
\nonumber \\
& = (x_1 x_2 \dots x_k ) \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{(n-1)/2}^2 + x_i^2) \big)
\nonumber
\end{align}

Now consider

\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k =
\nonumber \\
& = \det
\begin{pmatrix}
(x_1 \dots x_k)^{\frac{1}{k}} & 0 & 0 & \dots \\
0 & \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & 0 & \dots \\
0 & 0 & - \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & \dots \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\end{pmatrix}^k
\nonumber \\
& =
(x_1 x_2 \dots x_k ) \big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\dots
\big[ \theta_{(n-1)/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}

We easily have from (3) that

\begin{align}
& (x_1 x_2 \dots x_k ) \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{(n-1)/2}^2 + x_i^2) \big)
\geq
\nonumber \\
&
(x_1 x_2 \dots x_k ) \big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\dots
\big[ \theta_{(n-1)/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}

which establishes the main result for general odd values of $n$:

$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad \mathrm {for \; odd \;} n .$Completeting the proof of problem 1.

Thanks! I was worried I'd have to type up a solution if no one solved it by month end!

The solution looks about right. Proving the $n=2$ case is definitely the key unlocking the problem, which you tackled in your second spoiler. I'm a bit short on time right now but will take a closer look later on.

 

[StoneTemplePython]

I think I have solved problem 1:

I split the proof into the parts:

Part (a) A few facts about real skew symmetric matrices.
Part (b): Proof for $n$ even.
(i) Looking at case $n = 2$.
(ii) Proving a key inequality. This will prove case $n=2$
(iii) Proving case for general even $n$ (then easy).
Part (c) Case of odd $n$.
(i) Proving case for $n = 3$ (easy because of part (b)).
(iii) Proving case for general odd $n$ (easy because of part (b)).

I went through it fairly granularly and did not see any flaws.

A couple thoughts:

1.) If you are so inclined in your first spoiler, you may make use of rule

2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.

I would be happy to accept basic results about skew symmetric matrices' eigenvalues having zero real component by spectral theory. On the other hand, your workthrough may be more instructive for 3rd parties reading it who don’t know the spectral theory underlying it – so they both have merits.

2.) The key insight for this problem, in my view, is figuring out the $n = 2$ case. Everything can be built off of this. The other insight is relating it to $\text{GM} \leq \text{AM}$ in some way. I think you basically re-created Cauchy’s forward-backward induction proof for $\text{GM} \leq \text{AM}$, in Part (b) albeit for additivity not for vanilla $\text{GM} \leq \text{AM}$. Since we are at month end, I will share another much simpler idea, which is the fact that 'regular' $\text{GM} \leq \text{AM}$ implies this result.

my take is that in Part (B) (II) when you are seeking to prove:

$(\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k$

or equivalently

$\Big((\theta^2 + x_1^2) \dots (\theta^2 + x_k^2)\Big)^{1/k} \geq \theta^2 + (x_1^2 \dots x_k^2)^{1/k}$
multiply each side by

$\big(\theta^2\big)^{-1}$
(which is positive and doesn't change the inequality) and define
$z_i := \frac{x_i^2}{\theta^2} \gt 0$

The relationship is thus:

$\Big(\prod_{i=1}^k (1 + z_i)\Big)^{1/k}= \Big((1 + z_1) \dots (1 + z_k)\Big)^{1/k} \geq 1 + (z_1 \dots z_k)^{1/k} = \Big(\prod_{i=1}^k 1\Big)^{1/k} + \Big(\prod_{i=1}^k z_i\Big)^{1/k}$

which is true by the super additivity of the Geometric Mean (which incidentally was a past challenge problem, but since it is not this challenge problem I think it is fine to assume it is common knowledge to mathematicians).
- - - -
To consider the case of any eigenvalues equal to zero, we can verify that the inequality holds with equality, which we can chain onto the above.
- - - -
I have a soft spot for proving this via $2^r$ for $r = \{1, 2, 3, ...\}$ and then filling in the gaps. Really well done. Forward backward-induction is a very nice technique, but a lot of book-keeping!

 

[fresh_42]

Here is the solution to the last open problem #9.

For a given a real Lie algebra $\mathfrak{g}$, we define
$$
\mathfrak{A(g)} = \{\,\alpha \, : \,\mathfrak{g}\longrightarrow \mathfrak{g}\,\,: \,\,[\alpha(X),Y]=-[X,\alpha(Y)]\text{ for all }X,Y\in \mathfrak{g}\,\}\quad (1)
$$
The Lie algebra multiplication is defined by

• $(2)$ anti-commutativity: $[X,X]=0$
• $(3)$ Jacobi-identity: $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$

a) $\mathfrak{A(g)}\subseteq \mathfrak{gl}(g)$ is a Lie subalgebra in the Lie algebra of all linear transformations of $\mathfrak{g}$ with the commutator as Lie product $[\alpha, \beta]= \alpha \beta -\beta \alpha \quad (4)$ because
\begin{align*}
[[\alpha,\beta]X,Y]&\stackrel{(4)}{=}[\alpha \beta X,Y] - [\beta\alpha X,Y]\\
&\stackrel{(1)}{=}[X,\beta \alpha Y]-[X,\alpha \beta Y]\\
&\stackrel{(4)}{=}[X,[\beta,\alpha]Y]\\
&\stackrel{(2)}{=}-[X,[\alpha,\beta]Y]
\end{align*}
b) The smallest non Abelian Lie algebra $\mathfrak{g}$ with trivial center is $\mathfrak{g}=\langle X,Y\,: \,[X,Y]=Y\rangle\,.$ It's easy to verify $\mathfrak{A(g)} \cong \mathfrak{sl}(2,\mathbb{R})\,$, the Lie algebra of $2 \times 2$ matrices with trace zero.

$\mathfrak{g}=\mathfrak{B(sl(}2,\mathbb{R}))$ is the maximal solvable subalgebra of $\mathfrak{sl}(2,\mathbb{R})$, a so called Borel subalgebra.

c) To show that $\mathfrak{g} \rtimes \mathfrak{A(g)}$ is a semidirect product given by $$[X,\alpha]:=[\operatorname{ad}X,\alpha]=\operatorname{ad}X\,\alpha - \alpha\,\operatorname{ad}X\quad (5)$$ we have to show that this multiplication makes $\mathfrak{A}(g)$ an ideal in $\mathfrak{g} \rtimes \mathfrak{A(g)}$ and a $\mathfrak{g}-$module.
\begin{align*}
[[X,\alpha]Y,Z]&\stackrel{(5)}{=}[[X,\alpha Y],Z] - [\alpha[X,Y],Z]\\
&\stackrel{(3),(1)}{=}-[[\alpha Y,Z],X]-[[Z,X],\alpha Y]+[[X,Y],\alpha Z]\\
&\stackrel{(3),(1)}{=}[[Y,\alpha Z],X]+[\alpha[Z,X],Y]\\&-[[Y,\alpha Z],X]-[[\alpha Z,X],Y]\\
&\stackrel{(2)}{=}[Y,\alpha[X,Z]]-[Y,[X,\alpha Z]]\\
&\stackrel{(5)}{=}-[Y,[X,\alpha Z]]
\end{align*}
and $\mathfrak{A(g)}$ is an ideal in $\mathfrak{g} \rtimes \mathfrak{A(g)}$. It is also a $\mathfrak{g}-$module, because $\operatorname{ad}$ is a Lie algebra homomorphism $(6)$ and therefore
\begin{align*}
[[X,Y],\alpha]&\stackrel{(5)}{=}[\operatorname{ad}[X,Y],\alpha]\\
&\stackrel{(6)}{=}[[\operatorname{ad}X,\operatorname{ad}Y],\alpha]\\
&\stackrel{(3)}{=}-[[\operatorname{ad}Y,\alpha],\operatorname{ad}X]-[[\alpha,\operatorname{ad}X],\operatorname{ad}Y]\\
&\stackrel{(2)}{=}[\operatorname{ad}X,[\operatorname{ad}Y,\alpha]]-[\operatorname{ad}Y,[\operatorname{ad}X,\alpha]]\\
&\stackrel{(5)}{=} [X,[Y,\alpha]]-[Y,[X,\alpha]]
\end{align*}
d) For the last equation with $\alpha \in \mathfrak{A(g)}$ and $X,Y,Z \in \mathfrak{g}$
$$[\alpha(X),[Y,Z]]+[\alpha(Y),[Z,X]]+[\alpha(Z),[X,Y]] =0\quad (7)$$
we have
\begin{align*}
[\alpha(X),[Y,Z]]&\stackrel{(3)}{=}-[Y,[Z,\alpha(X)]]-[Z,[\alpha(X),Y]]\\
&\stackrel{(1)}{=} [Y,[\alpha(Z),X]]+[Z,[X,\alpha(Y)]]\\
&\stackrel{(3)}{=} -[\alpha(Z),[X,Y]]-[X,[Y,\alpha(Z)]]\\
&-[X,[\alpha(Y),Z]]-[\alpha(Y),[Z,X]]\\
&\stackrel{(1)}{=}-[\alpha(Y),[Z,X]]-[\alpha(Z),[X,Y]]
\end{align*}