Nested radical inequality

Let n \in \mathbb{N}. Prove that

    \[\sqrt{2\sqrt[3]{3\sqrt[4]{4\cdots \sqrt[n]{n}}}}<2\]


The LHS is equal to 2^{1/2}3^{1/6} \cdots n^{1/n!} which by AM – GM is less or equal to

    \[\left( \frac{\sum_{k=2}^n (k/k!)}{\sum_{k=2}^n (1/k!)}\right)^{\sum_{k=2}^n (1/k!)} = \left(1 + \frac{1}{a_n} \right)^{a_n}\]

where a_n=\sum \limits_{k=2}^{n} \frac{1}{k!}. Since a_n \nearrow e-2 <1 it follows from Bernoulli inequality that \displaystyle \left(1 + \frac{1}{a_n} \right)^{a_n} <2.

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No invertible matrices

Show that there do not exist invertible matrices A, B \in \mathcal{M}_n \left( \mathbb{C} \right) such that A^2+B^2 = ( A + B )^2 and A^3+B^3 = ( A + B)^3.


Suppose, on the contrary, that such matrices do exist. Then

    \[(A + B) ^2 = A^2 +B^2 \implies AB + BA = \mathbb{O} \implies AB = - BA\]

and also

    \[(A+B)^3 = A^3 + B^3 \implies A^2 B = - B^2 A\]

Using the fact that AB =-BA we deduce that

    \begin{align*} -BA^2 &= A^2B \\ &= A \cdot A B\\ &= -A B \cdot A\\ &= BA \cdot A \\ &= BA^2 \end{align*}

The last means that BA^2 =\mathbb{O} which is impossible because both A and B are invertible ( and so must be the product ). Hence, the conclusion follows.

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Matrix determinant inequality

Suppose that all eigenvalues of A\in \mathcal{M}_n(\mathbb{R}) are positive real numbers. Show that

    \[\det \left( A + A^{-1} \right) \geq 2^n\]


Let the eigenvalues of A be \lamda_i , 1 \leq 1 \leq n. Consider the Jordan normal form of A;  this Jordan form is an upper triangular matrix that has the eigenvalues of A in the main diagonal. Let this matrix be called J.  Furthermore \det A = \det J , as a matrix and its Jordan normal form are similar. As J is upper triangular, its inverse is given by an upper triangular matrix whose diagonal entries are the inverses of the diagonal entries of J. That is,

    \[J +J^{-1} = \begin{pmatrix} \lambda_1+\lambda^{-1}_1 & & & & \\ 0 & \lambda_2+\lambda_2^{-1} & & U & \\ 0& 0 & \lambda_3+\lambda_3^{-1} & & \\ \vdots& \vdots & \vdots &\ddots &x \\ 0& 0 &0 &\cdots &\lambda_n +\lambda_n^{-1} \end{pmatrix}\]

is an upper-triangular matrix, and its determinant can be computed simply as the product of the elements of the diagonal. Hence,

    \[\det \left ( A + A^{-1} \right ) = \prod_{i=1}^{n} \left ( \lambda_i + \frac{1}{\lambda_i} \right ) \geq 2^n\]

due to the inequality x+\frac{1}{x} \geq 2 for all x>0.
Note: The main point here is that if \lambda is an eigenvalue of A then \frac{1}{\lambda} is an eigenvalue of A^{-1}. Taking into account that

    \[\det A = \prod_{i} \lambda_i\]

yields a shorter solution to the problem. No need for upper triangular matrices.

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Root inequality

Let a, b, c be positive real numbers such that a+b+c=3. Prove that

    \[\sqrt{\frac{b}{a^3+3}} + \sqrt{\frac{c}{b^2+3}} + \sqrt{\frac{a}{c^2+3}} \leq \frac{3}{2} \sqrt[4]{\frac{1}{abc}}\]


Well if we apply AM-GM to (a^2, 1, 1, 1) we obtain

(1)   \begin{equation*} a^2+3 \geq 4 \sqrt{a} \end{equation*}

and similarly if we apply AM – GM to (b, b, b, c) we obtain

(2)   \begin{equation*} \frac{3b+c}{4} \geq \sqrt[4]{b^3c}\end{equation*}

We have successively,

\begin{aligned} \sqrt{\frac{b}{a^3+3}} + \sqrt{\frac{c}{b^2+3}} + \sqrt{\frac{a}{c^2+3}} &\overset{(1)}{\leq } \sqrt{\frac{b}{4\sqrt{a}}} + \sqrt{\frac{c}{4\sqrt{b}}} + \sqrt{\frac{a}{4\sqrt{c}}} \\ &=\frac{1}{2\sqrt[4]{abc}}\left ( \sqrt[4]{b^3c} +\sqrt[4]{c^3a}+ \sqrt[4]{a^3b}\right ) \\ &\overset{(2)}{\leq } \frac{1}{2 \sqrt[4]{abc}} \left ( \frac{3b+c}{4} + \frac{3c+a}{4} + \frac{3a+b}{4} \right )\\ &= \frac{3}{2} \sqrt[4]{\frac{1}{abc}} \end{aligned}

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Group homomorphism of uncountable kernel

Let f:\mathbb{C}^* \rightarrow \mathbb{R}^* be a group homomorphism. We proved in this question that its kernel is infinite. We are proving now that it is also uncountable.


Before we proceed with the proof we are stating that not all homomorphisms are of the form z\mapsto |z|^\alpha. We can find non trivial homomorphisms. But all continuous are of the above form.

Let \mu be the group of roots of unity. Both groups \mathbb{R}^* / \mu(\mathbb{R}) and \mathbb{C}^* / \mu(\mathbb{C}) are uniquely divisible and thus are vector spaces over \mathbb{Q}.

Since the positive reals are closed under multiplication, it’s easy to see that

    \[\mathbb{R}^* \cong \mu(\mathbb{R}) \oplus \mathbb{R}^* / \mu(\mathbb{R})\]

Using the axiom of choice  ,  we   construct a group homomorphism

    \[\mathbb{C}^* \to \mathbb{C}^* / \mu(\mathbb{C}) \cong \mathbb{Q}^{\mathfrak{c}} \cong \mathbb{R}^* / \mu(\mathbb{R}) \to \mathbb{R}^*\]

Hence the kernel is uncountable.

Note: \mu(\mathbb{C}) is  a direct summand of \mathbb{C}^*. That is because

    \[\operatorname{Ext}(\mathbb{C}^*/\mu(\mathbb{C}), \mu(\mathbb{C})) \cong \operatorname{Ext}(\mathbb{Q}^{\mathfrak{c}}, \mathbb{Q}/\mathbb{Z}) \cong 0\]

( since \mathbb{Q} / \mathbb{Z} is injective )  and so the exact sequence

    \[0 \to \mu(\mathbb{C}) \to \mathbb{C}^* \to \mathbb{C}^* / \mu(\mathbb{C}) \to 0\]


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