# Another Integral Inequality from the RMM

### Proof

For $a\ge 1,$

\displaystyle\begin{align} \Omega(a) &= \int_{-1/a}^{1/a}(2x^6+2x^4+3)\arccos(ax)dx\\ &=\int_{-1/a}^{1/a}(2x^6+2x^4+3)\left(\frac{\pi}{2}-\arcsin(ax)\right)dx\\ &=\frac{\pi}{2}I_1-I_2, \end{align}

where

\displaystyle\begin{align} I_1 &= \int_{-1/a}^{1/a}(2x^6+2x^4+3)dx\\ &=2\left[\frac{2}{7}x^7+\frac{2}{5}x^5+3x\right]_{0}^{1/a}\\ &=2\left[\frac{2}{7a^7}+\frac{2}{5a^5}+\frac{3}{a}\right]\\ &\le\frac{2}{35}[10+14+105]\frac{1}{a}\\ &=\frac{2\cdot 129}{35a}, \end{align}

because $a^7\ge a^5\ge a.\,$ On the other hand, $I_2=0,\,$ being an integral of an odd function. So, finally,

$\displaystyle\Omega(a)=\frac{\pi}{2}I_1-I_2\le\frac{\pi}{2}\cdot\frac{2\cdot 129}{35a}=\frac{129\pi}{35a}.$

### Acknowledgment

The problem from the Romanian Mathematical Magazine has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru, along with the solution by Ravi Prakash (India).

Copyright © 1996-2018 Alexander Bogomolny

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