An inequality

Let a, b, c be positive real numbers such that

a+b+c=1

Prove that

\displaystyle \prod_{\text{cyclic}}\left ( \frac{1}{a} + \frac{1}{bc} \right ) > 1728

(Vojtech Jarnik / Second Category / 2016)

Solution

Using the AM – GM inequality we have that

    \begin{align*} \frac{1}{a} + \frac{1}{bc}&= \frac{1}{a} + \frac{1}{3bc} + \frac{1}{3bc} + \frac{1}{3bc} \\ &> \frac{4}{\sqrt[4]{27 ab^3 c^3}} \end{align*}

as well as \displaystyle \left ( \frac{a+b+c}{3} \right )^3 = \frac{1}{27} >abc. Thus:

    \begin{align*} \prod_{\text{cyclic}} \left ( \frac{1}{a} + \frac{1}{bc} \right ) &> 64 \prod_{\text{cyclic}} \frac{1}{\sqrt[4]{27ab^3c^3}} \\ &=\frac{64}{\sqrt[4]{3^9 (abc)^7}} \\ &> \frac{64}{\sqrt[4]{3^9 \left ( 3^{-3} \right )^7}}\\ &= 64 \cdot \sqrt[4]{3^{12}} \\ &= 1728 \end{align*}

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Uniform convergence

Given the sequence of functions f_n:\mathbb{R} \rightarrow \mathbb{R} where n \in \mathbb{N} such that

\displaystyle f_n(x) = \frac{n}{n^3+x^2}

Prove that

(i) the serieses \displaystyle \sum_{n=1}^{\infty} f_n and \displaystyle \sum_{n=1}^{\infty} f'_n converge uniformly to functions f, g:\mathbb{R} \rightarrow \mathbb{R} .

(ii) the functions  f, g are continuous.

(iii) f'=g.

(iv) it holds that

\displaystyle \int_{-1}^{1} f(x) \, {\rm d}x = 2 \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \arctan \frac{1}{n \sqrt{n}} \quad, \quad \int_{-\pi}^{\pi} x^4 g(x) \, {\rm d}x = 0

Solution

(i) This is an immediate consequence of the Weierstrass M Test . We simply note that \displaystyle 0 \leq f_n(x)\leq \frac{1}{n^2} and of course

\displaystyle |f_n'(x)|=\frac{|2nx|}{(n^3+x^2)^2} \leq \frac{2}{n^{7/2}}

(ii) The uniform limit of continuous functions is continuous. This is enough for us to extract that both f and g throughout \mathbb{R}.

(iii) For a limit of differentiable functions, a sufficient condition for claiming that the derivative of the limit is the limit of the derivatives is the uniform convergence of the sequence of derivatives. Since the series for g converges uniformly, we can claim that f'=g.

(iv) For the first integral we have that

    \begin{align*} \int_{-1}^{1} f(x) \, {\rm d}x &= \int_{-1}^{1} \sum_{n=1}^{\infty} f_n(x) \, {\rm d}x \\ &= \sum_{n=1}^{\infty} \int_{-1}^{1} f_n(x) \, {\rm d}x\\ &= \sum_{n=1}^{\infty} n \int_{-1}^1\frac{{\rm d}x}{n^3+x^2}\\ &= 2\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \arctan \frac{1}{n \sqrt{n}} \end{align*}

and for the second integral we have

    \begin{align*} \int_{-\pi}^{\pi} x^4 g(x) \, {\rm d}x &=\int_{-\pi}^{\pi} x^4 \sum_{n=1}^{\infty} f_n'(x) \, {\rm d}x \\ &=2 \int_{-\pi}^{\pi} x^4 \sum_{n=1}^{\infty} \frac{nx}{\left ( n^3 +x^2 \right )^{2}} \, {\rm d}x \\ &= 2\sum_{n=1}^{\infty} n \cancelto{0}{\int_{-\pi}^{\pi} \frac{x^5}{\left ( n^3 +x^2 \right )^{2}} \, {\rm d}x}\\ &= 0 \end{align*}

since the function \displaystyle \frac{x^5}{\left ( n^3 +x^2 \right )^{2}} is odd.

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On matrices

Let A, B be two 3 \times 3 matrices with real entries. Prove that

A - \left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} = ABA

provided all the inverses appearing on the left hand side exist.

(Vojtech Jarnik / 2nd Category/ 2015)

Solution

Let A, B be elements of an arbitrary associative algebra with unit. Then:

\begin{aligned} \left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} &= \left ( A^{-1} \left ( B^{-1} - A \right )\left ( B^{-1} - A \right )^{-1} + A^{-1} A \left ( B^{-1} - A \right )^{-1} \right )^{-1} \\ &=\left ( A^{-1} \left ( \left ( B^{-1} - A \right ) +A \right )\left ( B^{-1} -A \right )^{-1} \right )^{-1} \\ &= \left ( A^{-1} B^{-1} \left ( B^{-1} -A \right )^{-1} \right )^{-1}\\ &= \left ( B^{-1}-A \right ) BA \\ &= A - ABA \end{aligned}

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A multiple sum

Evaluate

\displaystyle \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty} \cdots\sum_{n_k=1}^{\infty} \frac{1}{n_1 \cdot n_2 \cdots n_k \left ( n_1+n_2+\cdots+n_k+1 \right )}

(Vojtech Jarnik / 2nd Category/ 2011)

Solution

We are applying the classical trick, thus:

\begin{aligned} \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty} \cdots\sum_{n_k=1}^{\infty} \frac{1}{n_1 \cdot n_2 \cdots n_k \left ( n_1+n_2+\cdots+n_k+1 \right )} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty} \cdots\sum_{n_k=1}^{\infty} \frac{1}{n_1 \cdot n_2 \cdots n_k} \int_{0}^{1}x^{n_1+n_2+\cdots+n_k} \, {\rm d}x\\ &= \int_{0}^{1} \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty} \cdots\sum_{n_k=1}^{\infty} \frac{1}{n_1 \cdot n_2 \cdots n_k} x^{n_1+n_2+\cdots+n_k} \, {\rm d}x \\ &= \int_{0}^{1} \left ( -\ln (1-x) \right )^k \, {\rm d}x\\ &= k! \end{aligned}

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Riemann integrability

Let f:[0, 1] \rightarrow \mathbb{R} be defined as:

f(x)= \left\{\begin{matrix} 0&, &x \in [0,1]\cap \left ( \mathbb{R} \setminus \mathbb{Q} \right ) \\ x_n &, &x=q_n \in [0,1] \cap \mathbb{Q} \\ \end{matrix}\right.

where x_n is a sequence such that \lim x_n =0 and 0 \leq x_n \leq 1 and q_n be an enumeration of the rationals of the interval [0, 1]. Prove that f is Riemann integrable and that \bigintsss_0^1 f(x)\, {\rm d}x=0.

Solution

We define

f_n = \left\{\begin{matrix} x_n &, & x \in \{ q_1, \dots, q_n\}\\ 0 & , & x \notin \{ q_1, \dots, q_n \} \end{matrix}\right.

Apparently f_n are Riemann integrable and \bigintsss_0^1 f_n(x) \, {\rm d}x =0. Also f_n \rightarrow f uniformly because f_n, f coincide except of some points q_m (m>n). Hence , since \lim x_n=0 , we have that:

0\leq \left | f(x)-f_n(x) \right |\leq \sup \left \{ x_m \mid m>n \right \} \xrightarrow{n \rightarrow +\infty}0

Thefore f is Riemann integrable and thus:

    \begin{align*} \int_{0}^{1} f(x) \, {\rm d}x &=\int_{0}^{1} \lim_{n \rightarrow +\infty} f_n (x) \, {\rm d}x \\ &= \lim_{n \rightarrow +\infty} \int_{0}^{1} f_n (x) \, {\rm d}x\\ &= \lim_{n \rightarrow +\infty}0 \\ &= 0 \end{align*}

The exercise can also be found in mathematica.gr

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