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Homogeneity of inequality

Let x, y, z>0. Prove that:

    \[\frac{y^3z}{x^2(xy+z^2)} +\frac{z^3x}{y^2(zy+x^2)} +\frac{x^3y}{z^2(xz+y^2)} \geq \frac{3}{2}\]

Solution

Due to homogeneity we may assume xyz=1. Thus there exist positive a, b, c such that

    \[x=\frac{a}{b}\quad , \quad y=\frac{b}{c}\quad ,\quad z=\frac{c}{a}\]

Hence,

    \begin{align*} \sum \frac{y^3z}{x^2(xy+z^2)}  &= \frac{a^5}{bc(b^3+c^3)}+\frac{b^5}{ca(c^3+a^3)}+\\ & \quad \quad \quad +\frac{c^5}{ab(a^3+b^3)}\\ &=\frac{a^6}{abc(b^3+c^3)}+\frac{b^6}{abc(c^3+a^3)}+\\ &\quad \quad \quad + \frac{c^6}{abc(a^3+b^3)} \\ &\geq \frac{(a^3+b^3+c^3)^2}{2abc(a^3+b^3+c^3)}\\ &=\frac{1}{2}\frac{a^3+b^3+c^3}{abc}\\ &\geq \frac{3}{2} \end{align*}

 

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Galois theory … of the Euler’s totient function

Let n>2 and let \omega \in \mathbb{C} be an n-th primitive root of unity. Prove that

    \[[\mathbb{Q}(\omega + \omega^{-1}) :\mathbb{Q}]=\frac{\varphi (n)}{2}\]

where \phi denotes the Euler’s totient function.

Solution

We have

    \[\phi(n)=[\mathbb{Q}(\omega):\mathbb{Q}]=[\mathbb{Q}(\omega):\mathbb{Q}(\omega +\omega^{-1})][\mathbb{Q}(\omega +\omega^{-1}):\mathbb{Q}]\]

where [\mathbb{Q}(\omega):\mathbb{Q}(\omega +\omega^{-1})]=2 since t^2-(\omega+\omega^{-1})t+1 is the minimal polynomial of \omega over \mathbb{Q}(\omega+\omega^{-1}).

The result follows.

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Complex inequality

Prove that forall z \in \mathbb{C} and n \in \mathbb{N} it holds that

    \[\left | \left ( 1+z \right )^n -1 \right | \leq \left ( 1+\left | z \right | \right )^n -1\]

Solution

Well,

    \begin{align*} \left | \left ( 1+z \right )^n -1 \right | &= \left | \sum_{k=0}^{n} \binom{n}{k} z^k -1 \right | \\ &= \left | \sum_{k=1}^{n} \binom{n}{k} z^k\right |\\ &\leq \sum_{k=1}^{n} \binom{n}{k} \left | z^k \right | \\ &=\sum_{k=0}^{n} \binom{n}{k} \left | z \right |^n -1 \\ &= \left ( 1+\left | z \right | \right )^n -1 \end{align*}

Done!

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Bound of derivative

Let \displaystyle f(x) = \frac{\sin x}{x} \; , \; x \neq 0. Let f^{(n)} denote the n-th derivative. Prove that

    \[\left | f^{(n)}(x) \right | < \frac{1}{n+1}\]

Solution

We note that

    \[\frac{\sin x}{x} = \int_{0}^{1} \cos \alpha x \, \mathrm{d}\alpha\]

Hence,

    \begin{align*} \left | f^{(n)}(x) \right | &= \left | (-1)^n \int_{0}^{1} \alpha^n \cos \alpha x \, \mathrm{d} \alpha \right | \\ &=\left | \int_{0}^{1} \alpha^n \cos \alpha x \, \mathrm{d} \alpha \right | \\ &< \int_{0}^{1} \alpha^n \, \mathrm{d} \alpha\\ &= \frac{1}{n+1} \end{align*}

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Searching for the …function

Find all \mathcal{C}^1 functions f:[0, 1] \rightarrow (0, +\infty) such that \frac{f(1)}{f(0)} = e and

    \[\int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \leq 2\]

Solution

First of all we note that

    \[\int_{0}^{1} \frac{f'(x)}{f(x)} \, \mathrm{d}x = \left [ \ln f(x) \right ]_0^1 = 1\]

Thus,

    \begin{align*} \int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \leq 2 &\Leftrightarrow \int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \\ & \quad \quad \quad \leq 2 \int_{0}^{1} \frac{f'(x)}{f(x)} \, \mathrm{d}x\\ &\Leftrightarrow \int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} - 2 \int_{0}^{1} \frac{f'(x)}{f(x)} \, \mathrm{d}x + \\ & \quad \quad \quad + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \leq 2 \\ &\Leftrightarrow \int_{0}^{1} \left ( \frac{1}{f(x)} - f'(x) \right )^2 \, \mathrm{d}x \leq 0 \end{align*}

Since the integrand is positive it only remains that

    \begin{align*} \frac{1}{f(x)}- f'(x) =0 &\Leftrightarrow \frac{1}{f(x)} = f'(x) \\ &\Leftrightarrow f'(x) f(x) = 1 \\ &\Leftrightarrow f^2(x) = 2x+ c \end{align*}

Setting x=0 and x=1 at the last equation we have that:

    \[\begin{matrix} f^2(0) & = & c \\ f^2(1) &= &2+c \end{matrix}\Rightarrow \left ( \frac{f(0)}{f(e)} \right )^2 = \frac{c}{2+c} \Rightarrow \frac{1}{e^2} = \frac{c}{2+c} \Rightarrow c = \frac{2}{e^2-1}\]

Since f is positive we conclude that

    \begin{align*} f^2(x) = 2x+ \frac{2}{e^2-1} &\Leftrightarrow f(x) = \sqrt{2x + \frac{2}{e^2-1}} \; , \; x \in [0, 1] \end{align*}

which satisfies the given conditions.

 

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