Inequality in acute triangle

Let ABC be an acute triangle. Prove the following inequality

    \[\sum \cot^2 {\rm A} \cot^2 {\rm B} \geq \frac{\sum \cos^2 {\rm A}}{\sum \sin^2 {\rm A}}\]

Solution

The solution can be found at cut -the – knot. 

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Inequality in a triangle

Given a triangle ABC let m_a , m_b , m_c denote the median points of the sides a , b, c respectively. Prove that

    \[\left ( m_a + m_b + m_c \right )^2 + 4 r \left ( R - 2r \right ) \leq \left ( 4 R + r \right )^2\]

where R denotes the the circumradius and r the inradius respectively.

(Adil Abdullayev / RMM)

Solution [Soumava Chakraborty]

We have successively

    \begin{align*} \left ( \sum m_a \right )^2 &=\sum m_a^2 + 2 \sum m_a m_b \\ &=\frac{3}{4} \sum a^2 + 2 \sum m_a m_b \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{m_am_b \leq \frac{2c^2+ab}{4}}{\leq } \frac{3}{4} \sum a^2 + \frac{1}{2} \sum \left ( 2c^2 + ab \right )\\ &= \frac{3}{4} \sum a^2 + \sum a^2 + \frac{1}{2} \sum ab\\ &= \frac{7}{4} \sum a^2 + \frac{1}{2} \sum ab \\ &= \frac{1}{4} \left [ \sum a^2 + 2 \sum ab + 6 \sum a^2 \right ]\\ &= \frac{4s^2 + 12 \left ( s^2-4Rr -r^2 \right )}{4} \\ &=4s^2 -12 Rr -3r^2 \\ &\!\!\!\!\!\!\!\!\!\overset{\text{Gerretsen}}{\leq } 4 \left ( 4R^2 + 4Rr + 3r^2 \right ) -12 Rr -3r^2 \\ &=16R^2 + 8Rr + r^2 -4Rr + 8r^2 \\ &=\left ( 4R+r \right )^2 - 4r \left ( R - 2r \right ) \end{align*}

In all the above s denotes the semiperimeter of the triangle. More on Gerretsen’s inequality can be found at this link .

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Inequality

Let x, y,z >0 satisfying x+y+z=1. Prove that

    \[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}\]

Solution

Since \frac{1}{x} \; , \; \frac{1}{y} \; , \; \frac{1}{z} >0 then the numbers

    \[\sqrt{\frac{1}{x} + \frac{1}{y}} \; , \; \sqrt{\frac{1}{x} +\frac{1}{z}} \; , \; \sqrt{\frac{1}{y} + \frac{1}{z}}\]

could be sides of a triangle. The area of this triangle is

    \[\mathcal{A} = \frac{1}{2} \sqrt{\frac{1}{xy} + \frac{1}{xz} + \frac{1}{yz}} = \frac{1}{2} \sqrt{\frac{x+z+y}{xyz}} = \frac{1}{2\sqrt{xyz}}\]

However , in any triangle is holds that [Weitzenböck]

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

where \mathcal{A} is the area of the triangle. Thus

    \[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}\]

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Exponential matrix

Let A \in \mathcal{M}_{n \times n} \left( \mathbb{C} \right). We define

    \[e^A = \sum_{m=0}^{\infty} \frac{A^m}{m!}\]

It is known that this series converges. Prove that

    \[\det (e^A) = e^{\Tr (A)}\]

Solution

We triangulise the matrix A , that is A= P^{-1} T P where P is an invertible matrix and T is an upper triangular. This is possible since our matrix is over \mathbb{C} and thus its characteristic polynomial splits. Let \lambda_1, \lambda_2, \dots, \lambda_n be its eigenvalues. Then we note that T^k is upper triangular with \lambda_1^k , \lambda_2^k , \dots, \lambda_n^k in its diagonal. Hence e^T is also upper triangular with e^{\lambda_1}, e^{\lambda_2} , \dots, e^{\lambda_n} in its diagonal. Hence

    \[\det e^T=e^{\lambda_1} e^{\lambda_2} \cdots e^{\lambda_n}=e^{\lambda_1+\lambda_2 +\cdots+\lambda_n}=e^{\Tr (T)}\]

However \Tr (A) = \Tr (T) and P^{-1} T^k P = A^k forall k. Thus P^{-1}e^TP=e^A and finally

    \[\det e^T=\det (P^{-1}e^TP)=\det e^A \]

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An arccot sum

Evaluate the sum

    \[\mathcal{S} = \sum_{n=1}^{\infty} \left( n \arccot n -1 \right)\]

Solution

Well, first of all we note that

    \begin{align*} n \arccot n - 1 &= n \arctan \left ( \frac{1}{n} \right ) -1 \\ &=n \int_{0}^{1/n} \frac{{\rm d}x}{1+x^2} -1 \\ &= -n \int_{0}^{1/n} \frac{x^2}{1+x^2}\, {\rm d}x \end{align*}

We also recall the Fourier series expansion of \pi \coth \pi z which is no other than

(1)   \begin{equation*} \pi \coth \pi z = \frac{1}{z} + \sum_{n=1}^{\infty} \frac{2z}{n^2+z^2} \end{equation*}

Hence

    \begin{align*} \mathcal{S} &= -\sum_{n=1}^{\infty} n \int_{0}^{1/n} \frac{x^2}{1+x^2} \, {\rm d}x\\ &=-\sum_{n=1}^{\infty} \int_{0}^{1} \frac{x^2}{x^2+n^2} \, {\rm d}x \\ &= \frac{1}{2} \int_{0}^{1} \left ( 1 - \pi x \coth \pi x \right ) \, {\rm d}x\\ &= \frac{1}{2} \int_{0}^{1} \left [ 1- \pi x \left ( 1 + \frac{2e^{-2\pi x}}{1-e^{-2\pi x}} \right ) \right ] \, {\rm d}x \\ &= \frac{1}{2} + \frac{17 \pi}{24} - \frac{1}{2} \log \left ( e^{2\pi} - 1 \right ) + \frac{1}{4\pi} {\rm Li}_2 \left ( e^{-2\pi} \right ) \end{align*}

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