# An Inequality in Triangle for Side Lengths, Cycled in Two Ways

### Problem

### Solution

By what's known as *Ravi's substitution*, there are $x,y,z \gt 0$ (the tangents from the vertices to the incircle) such that $a=y+z,$ $b=z+x,$ and $c=x+y$ (all $x,y,z$ are positive.) Rewrite the inequality as

$\displaystyle 3\sum_{cycl}\frac{x+y}{y+z}-2\sum_{cycl}\frac{y+z}{x+y}-3\ge 0.$

Getting rid of the denominators leads to a cyclic homogeneous $3^{rd}$ degree polynomial inequality $f(x,y,z)\ge 0$ which may suggest employing Sladjan Stankovick's CD3 theorem.

Obviously, $f(1,1,1)=0\ge 0.$ Due to the homogeneity of $f$ suffice it to check that $f(x,1,0)\ge 0$ which is $x^3+2x^2-3x+1\ge 0.$ The latter is equivalent to

$x(1-x)(x+3)\le 1.$

Now observe that for $x\in [0,1],$ $\displaystyle 0\le x(1-x)\le\frac{1}{4},$ whereas $x+3\le 4,$ implying that, for $x\in[0,1],$ $x(1-x)(x+3)\le 1.$ For, $x\ge 1,$ $x(1-x)(x+3)\le 0\lt 1.$ It follows that $x(1-x)(x+3)\le 1$ for all $x\ge 0.$

### Acknowledgment

The problem was been kindly posted by Leo Giugiuc at the CutTheKnotMath facebook page, with a solution of his.

[an error occurred while processing this directive]

|Contact| |Up| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny