An Inequality with Integrals and Rearrangement

Problem

Leo Giugiuc has kindly communicated to me the following problem, along with a solution. The problem is from Dan Sitaru's book Math Accent.

An Inequality with Integrals and Rearrangement

Solution

First off, $\displaystyle\lim_{x\rightarrow0^+}(\cot x\cdot\arctan x)=\lim_{x\rightarrow0^+}\left(\cos x\cdot\frac{\arctan x}{\sin x}\right)=1,\;$ implying that the function $\displaystyle f:\;[0,\pi)\rightarrow\mathbb{R},\;$ defined by

$f(x)=\begin{cases} \cot x\cdot\arctan x, & x\in(0,\pi)\\ 1,& x= 0, \end{cases}$

is continuous.

Now, for $\displaystyle x\in\left(0,\frac{\pi}{2}\right),\;$ $\arctan x\lt x\lt \tan x,\;$ implying that $\cot x\cdot\arctan x\lt\cot x\cdot\tan x=1.\;$ It follows that on $\displaystyle \left[0,\frac{\pi}{2}\right],\;$ $f(x)\lt 1\;$ so that $\displaystyle\int_{0}^{a}f(x)dx\lt\int_{0}^{a}1dx\lt a,\;$ if $\displaystyle a\in\left(0,\frac{\pi}{2}\right].$

An Inequality with Integrals and Rearrangement, illustration

If $\displaystyle a\in\left(\frac{\pi}{2},\pi\right)\;$ then

$\displaystyle\int_{0}^{a}f(x)dx=\int_{0}^{\pi/2}f(x)dx+\int_{\pi/2}^{a}f(x)dx\lt\int_{0}^{\pi/2}f(x)dx\lt\frac{\pi}{2}\lt a.$

Thus, for $a\in (0,\pi),\;$ $\displaystyle\int_{0}^{a}f(x)dx\lt a\;$ and similarly $\displaystyle\int_{0}^{b}f(x)dx\lt b\;$ and $\displaystyle\int_{0}^{c}f(x)dx\lt c.\;$ Thus, $\displaystyle\sum_{cycl}b^2c^3\int_{a}^af(x)dx\lt\sum_{cycl}ab^2c^3.\;$ But $\displaystyle\sum_{cycl}ab^2c^3=abc\sum_{cycl}bc^2\;$ and, by the rearrangement inequality, $\displaystyle\sum_{cycl}bc^2\;\le a^3+b^3+c^3.$

 

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