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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Sinc series

The sinc function is defined as:

    \[\mathrm{sinc} \; x =\left\{\begin{matrix} 1 & , & x=0 \\\\ \dfrac{\sin x}{x} &, & x \neq 0 \end{matrix}\right.\]

Prove that for any couple (\alpha, \beta) of real numbers in (0, 1) the following result holds:

    \[\sum_{n=-\infty}^{\infty} \mathrm{sinc} \; na \; \mathrm{sinc} \; n \beta = \frac{\pi}{\max\{\alpha, \beta \}}\]

Solution

Since

    \[\sum_{n=-\infty}^{\infty} \mathrm{sinc} \; na \; \mathrm{sinc} \; n \beta = 1 + \frac{2}{\alpha \beta} \sum_{n=1}^{\infty} \frac{\sin n \alpha \sin n \beta}{n^2}\]

due to the addition formulas for the sine and cosine functions it is enough to prove the equality

    \[f(\theta) = \sum_{n=1}^{\infty} \frac{\cos n \theta}{n^2} = \frac{\pi^2}{6} - \frac{\theta\left ( 2 \pi - \theta \right )}{4}  \quad \text{forall} \quad \theta \in [0, 2\pi]\]

which is an immediate consequence of the Fourier series by integrating the sawtooth wave function;

    \[\sum_{n=1}^{\infty} \frac{\sin n \theta}{n} = \frac{\pi-\theta}{2}\]

Hence,

    \begin{align*} \sum_{n=-\infty}^{\infty} \mathrm{sinc} \; na \; \mathrm{sinc} \; n \beta &= 1+ \frac{f\left ( \left | \alpha- \beta\right | \right) + f\left ( \alpha+\beta \right )}{\alpha \beta} \\ &=1+\frac{1}{4\alpha \beta} \bigg ( \left ( \alpha-\beta \right )^2 - \left ( \alpha+\beta \right )^2 +\\ &\quad \quad \quad + 2\pi \left ( \alpha-\beta-\left | \alpha+\beta \right | \right ) \bigg ) \\ &= 1+ \frac{1}{4\alpha \beta} \left ( -4\alpha \beta + 4 \pi \min \{ \alpha, \beta \} \right )\\ &= \frac{\pi}{\max\{\alpha, \beta \}} \end{align*}

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Continuous binomial integral

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

    \[\int_{-\infty}^{\infty} \binom{n}{x} \, \mathrm{d}x = 2^n\]

Solution

First of all,

    \begin{align*} \binom{n}{x} &= \frac{n!}{x! \left ( n-x \right )!} \\ &=\frac{n!}{\Gamma(x+1) \Gamma(n+1-x)} \\ &=\frac{n!}{\Gamma(1+x)(n-x)\left ( n-1-x \right )\cdots (1-x) \cdot \Gamma(1-x)} \\ &= \frac{n!}{x \Gamma(x)(n-x)\left ( n-1-x \right )\cdots (1-x) \Gamma(1-x) } \\ &= \frac{n!}{\pi} \cdot \frac{\sin \pi x}{(n-x)\left ( n-1-x \right )\cdots (1-x) \cdot x} \end{align*}

due to the well known formulae \Gamma(x+1) = x \Gamma(x) and \Gamma(x) \Gamma(1-x )  = \pi \csc \pi x.

However using the residue theorem we get that

    \[\frac{1}{(n-x)\left ( n-1-x \right )\cdots (1-x) \cdot x} = \sum_{k=0}^{n} \frac{1}{x-k} \cdot \frac{(-1)^k}{n!} \binom{n}{k}\]

Thus,

    \begin{align*} \int_{-\infty}^{\infty} \binom{n}{x} \, \mathrm{d}x &= \int_{-\infty}^{\infty} \frac{n!}{\pi} \cdot \frac{\sin \pi x}{(n-x)\left ( n-1-x \right )\cdots (1-x) \cdot x} \, \mathrm{d}x \\ &= \int_{-\infty}^{\infty} \frac{n!}{\pi} \cdot \sum_{k=0}^{n} \frac{\sin \pi x}{x-k} \cdot \frac{(-1)^k}{n!} \binom{n}{k} \, \mathrm{d}x \\ &=\frac{1}{\pi}\sum_{k=0}^{n} (-1)^k \binom{n}{k}\int_{-\infty}^{\infty} \frac{\sin \pi x}{x-k} \, \mathrm{d}x \\ &=\frac{1}{\pi} \sum_{k=0}^{n} (-1)^k \binom{n}{k} (-1)^k \pi \\ &= \sum_{k=0}^{n} \binom{n}{k} \\ &=2^n \end{align*}

due to the binomial theorem and the well known fact

    \[\int_{-\infty}^{\infty} \frac{\sin \pi x}{x-k} \, \mathrm{d}x = (-1)^k \pi\]

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Trigonometric inequality on a triangle

Let ABC be a triangle. Prove that

    \[\frac{1}{\sin^2 A} + \frac{1}{\sin^2 B} + \frac{1}{\sin^2 C} \geq 4\]

Solution

The function \displaystyle f(x) = \frac{1}{\sin^2 x } is convex, thus:

    \begin{align*} \frac{1}{3} \sum f(A) &= \frac{1}{3} \sum \frac{1}{\sin^2 A}\\ &\geq f\left ( \frac{1}{3} \sum A \right ) \\ &= \frac{1}{\sin^2 \left ( \frac{1}{3} \sum A \right )} \\ &= \frac{1}{\sin^2 \frac{\pi}{3}} \\ &= \frac{4}{3} \end{align*}

The result follows. In fact, something a bit stronger holds. Let r denote the inradius , R the circumradius and s the semiperimeter. Then,

    \[\frac{1} {\sin ^2 A} + \frac{1} {\sin^2 B} + \frac{1} {\sin^2 C} \geq \frac{2R}{r}\]

Indeed,

    \begin{align*} \sum \frac{1}{\sin^2 A} &= 4R^2\sum \frac{1}{a^2} \\ &\geq 4R^2 \sum \frac{1}{ab}\\ &=\frac{4R^2 \left ( a+b+c \right )}{abc}\\ &= \frac{8R^2s}{abc} \\ &= 8R^2 \cdot \frac{{\rm E}}{r} \cdot \frac{1}{4R{\rm E}} \\ &= \frac{2R}{r} \end{align*}

in view of the known identities

(1)   \begin{equation*} E = rs \end{equation*}

(2)   \begin{equation*} 4R\mathrm{E} = abc \end{equation*}

(3)   \begin{equation*} R \geq 2 r \end{equation*}

 

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A trigonometric limit

Evaluate the limit:

    \[\ell = \lim_{n \rightarrow +\infty} \int_0^1 \frac{|\sin nx|}{x^2+1}\, \mathrm{d}x\]

Solution

Let us begin with the Fourier series of |\sin x | which is of the form:

    \[|\sin x | = \frac{2}{\pi} + \sum_{n=1}^{\infty} a_n \cos nx\]

Hence

(1)   \begin{equation*} |\sin n x | = \frac{2}{\pi} + \sum_{m=1}^{\infty} a_m \cos n m x  \end{equation*}

Integrating ( 1 ) we get that:

    \begin{align*} \ell &=\lim_{n \rightarrow +\infty} \int_{0}^{1} \frac{|\sin n x|}{1+x^2} \, \mathrm{d}x \\ &= \lim_{n \rightarrow +\infty} \int_{0}^{1} \frac{1}{x^2+1} \left ( \frac{2}{\pi} + \sum_{m=1}^{\infty} a_m \cos nm x \right ) \, \mathrm{d}x\\ &=\lim_{n \rightarrow +\infty} \left ( \frac{2}{\pi} \int_{0}^{1} \frac{\mathrm{d}x}{x^2+1} + \int_{0}^{1} \frac{1}{x^2+1} \sum_{m=1}^{\infty} a_m \cos nm x \, \mathrm{d}x \right ) \\ &= \frac{1}{2} + \lim_{n \rightarrow +\infty } \sum_{m=1}^{\infty} a_m  \int_{0}^{1}  \frac{\cos nmx}{x^2+1} \, \mathrm{d}x \end{align*}

The Riemann – Lebesgue Lemma implies that

    \[\lim_{n \rightarrow +\infty} \int_{0}^{1}  \frac{\cos nmx}{x^2+1} \, \mathrm{d}x  =0\]

and thus \ell=\frac{1}{2}.

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