# An Easy Inequality with Three Integrals

### Problem

### Remark

The starting point of the solutions below is the observation that $\displaystyle\frac{\arctan x}{x}\lt 1,\;$ for all $x\gt 0.\;$ The fraction has a limit of 1 as $x\rightarrow 0^{+}\;$ which allows for the definition (by continuity) $f(1)=1.\;$

The inequality is equivalent to $\displaystyle\frac{\theta}{\tan\theta}\lt 1,\;$ for $\displaystyle\theta\in\left(0,\frac{\pi}{2}\right).$

Obviously, the inequality holds for any $f(x)\le 1\;$ in leau of $\displaystyle\frac{\arctan x}{x}.$

### Solution 1

We have

$\displaystyle \int_{0}^{u}\frac{\arctan x}{x}dx \lt 1.$

It follows that

$\displaystyle a^2\int_{0}^{b}\frac{\arctan x}{x}dx + b^2\int_{0}^{c}\frac{\arctan x}{x}dx + c^2\int_{0}^{a}\frac{\arctan x}{x}dx \lt a^2b+b^2c+c^2a.$

Suffice it to show that $a^2b+b^2c+c^2a\le a^3+b^3+c^3.\;$ By the AM-GM inequality,

$a^3+a^3+b^3\ge 3\sqrt[3]{a^6b^3}=3a^2b,\\ b^3+b^3+c^3\ge 3b^2c,\\ c^3+c^3+a^3\ge 3c^2a.$

Summing up gives $3(a^2+b^3+c^3)\ge 3(a^2b+b^2c+c^2a),\;$ as desired.

### Solution 2

This solution only differs from the above in treatment of $a^2b+b^2c+c^2a\le a^3+b^3+c^3.\;$ This is simply true by the rearrangement inequality.

### Acknowledgment

The problem above has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru, along with a solution (Solution 1) by Soumava Chakraborty. Leo Giugiuc and Ravi Prakash have commented with practically identical solutions (Solution 2). The inequality has been published in the Romanian Mathematical Magazine.

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