# A Cyclic Inequality in Three Variables IX

### Solution 1

Let $\displaystyle a=\frac{x}{y},\,$ $\displaystyle b=\frac{y}{z},\,$ $\displaystyle a=\frac{z}{x}.\,$ Note that $abc=1.\,$ The given inequality rewrites as

$\displaystyle 9(a^2+b^2+c^2)^2\ge 8(a+b+c)(a^3+b^3+c^3-3abc).$

Using $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\quad$ and rearranging, this reduces to

$\displaystyle \left(\sum_{cycl}a^2\right)^2-8\left(\sum_{cycl}ab\right)\left(\sum_{cycl}a^2\right)+16\left(\sum_{cycl}ab\right)^2\ge 0,$

or, $(a^2+b^2+c^2-4ab-4bc-4ca)^2\ge 0\,\,$ which is obviously true.

### Solution 2

Set $\displaystyle f=9\sum_{cycl}\frac{x^2}{y^2}-8\left(\sum_{cycl}\frac{x}{y}\right)\left(-3 \sum_{cycl}\frac{x^3}{y^3}\right).\,$ We need to prove that $f\ge 0.\,$ Factoring we get

$\displaystyle f = \frac{\displaystyle 9\left(\sum_{cycl}x^2y^4\right)^2}{x^4y^4z^4} -\frac{\displaystyle 8\left(\sum_{cycl}xy^2\right)^2\left(\sum_{cycl}x^2y^4-\sum_{cycl}x^3y^2z\right)}{x^4y^4z^4}.$

The numerator reduces to $\displaystyle\left(\sum_{cycl}x^2y^4-4\sum_{cycl}x^3y^2z\right)^2\ge 0.$

### Acknowledgment

Dan Sitaru has kindly posted the above problem (from his book "Math Accent") at the CutTheKnotMath facebook page, along with a solution (Solution 1) by Saptak Bhattacharya. Solution 2 is by Nassim Nicholas Taleb. The illustration is by Gary Davis.