# A 4-variable Inequality from the RMM

### Problem

### Proof

Denote $x=b+c+\eta,\;$ $y=c+a+\eta,\;$ $z=a+b+\eta.\;$ Then, e.g., $a-b=y-x,\;$ and the inequality to prove becomes

$\displaystyle |(x-y)(y-z)(z-x)|\le\sum_{cycl}|(x-y)xy|.$

We have,

$\displaystyle\begin{align} RHS &= \sum_{cycl}|(x-y)xy|\\ &\ge |(x-y)xy+(y-z)yz+(z-x)zx|\\ &= |(x-y)(y-z)(z-x)|\\ &= |(a-b)(b-c)(c-a)|, \end{align}$

as required. Equality is achieved when $x=y=z,\;$ i.e., when $a=b=c.$

### Acknowledgment

Dan Sitaru has kindly posted the above problem from the Romanian Mathematical Magazine, with a proof by Ravi Prakash, at the CutTheKnotMath facebook page.

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