# An Inequality in Fractions with Absolute Values

### Solution

Note that function $f(x,y)=\displaystyle \frac{x|x|-y|y|}{x-y}\,$ satisfy $f(x,y)=f(-x,-y),\,$ and so too

$f(-x,-y)+f(-y,-z)+f(-z,-x)=f(x,y)+f(y,z)+f(z,x).$

If so, suffice it to consider only two cases: 1) $a,b,c\ge 0\,$ and 2) $a,b\ge 0\,$ and $c\le 0.\,$

In the first case, we can simply remove the absolute values throughout to obtain

\displaystyle \begin{align} \min\{(a+b),(b+c),(c+a)\} &\lt \frac{1}{3}\sum_{cycl}\frac{a^2-b^2}{a-b}\\ &=\frac{1}{3}\sum_{cycl}(a+b)\\ &\lt 2\max\{a,b,c\}. \end{align}

In the second case, let $|c|=-c.\,$ Then

\displaystyle\begin{align} f(a,b)+f(b,c)+f(c,a)&=\frac{a|a|-b|b|}{a-b}+\frac{b|b|-c|c|}{b-c}+\frac{c|c|-a|a|}{c-a}\\ &=(a+b)+\frac{b^2+|c|^2}{b+|c|}+\frac{|c|^2+a^2}{|c|+a}\\ &\lt (a+b)+(b+|c|)+(|c|+a)\\ &\lt 6\max\{|a|,|b|,|c|\}. \end{align}

On the other hand, for $x,y\ge 0,\,$ $x^2+y^2\ge |x^2-y^2|=|x-y|\cdot (x+y),\,$ with equality only when of $x,y\,$ is zero. Thus, it follows that $\displaystyle\frac{b^2+|c|^2}{b+|c|}\ge |b-|c||=|b+c|\,$ and, similarly, $\displaystyle\frac{|c|^2+a^2}{|c|+a}\ge |a-|c||=|c+a|.\,$ Adding up gives

\displaystyle\begin{align} \frac{a|a|-b|b|}{a-b}+\frac{b|b|-c|c|}{b-c}+\frac{c|c|-a|a|}{c-a}&\ge (a+b)+|b+c|+|c+a|\\ &\gt 3\min\{|a+b|,|b+c|,|c+a|\} \end{align}

No two of $a,b,c\,$ may vanish simultaneously.

### Acknowledgment

This problem from his book "Algebraic Phenomenon" has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru.

Soumava Chakraborty gave a proof of $\displaystyle |x+y|\le\frac{x|x|-y|y|}{x-y}\le |x|+|y|\,$ by considering four cases and a similar proof has been submitted by Ravi Prakash.