# An Inequality with Cyclic Sums on Both Sides III

### Solution

By the AM-GM inequality,

$x^6z^3+x^6z^3+y^6x^3 \ge 3\sqrt[3]{x^{15}z^6y^6}=3x^5y^2z^2.$

Similarly,

\displaystyle \begin{align} &y^6x^3+y^6x^3+z^6y^3\ge 3x^2y^5z^2\\ &z^6y^3+z^6y^3+x^6z^3\geq 3x^2y^2z^5 \end{align}

$3(x^6z^3+y^6x^3+z^6y^3)\geq 3x^2y^2z^2(x^3+y^3+z^3),$

i.e.,

$x^6z^3+y^6z^3+z^6y^3\geq x^2y^2z^2(x^3+y^3+z^3)$

Suffice it to prove that

$\displaystyle \frac{x^2y^2z^2(x^3+y^3+z^3)}{x^2y^2z^2}\geq \frac{x^3+y^3+z^3+3xyz}{2},$

or, equivalently,

$\displaystyle 2(x^3+y^3+z^3)\geq x^3+y^3+z^3+3xyz,$

i.e.,

$\displaystyle x^3+y^3+z^3\geq 3xyz,$

which is true by the AM-GM inequality: $x^3+y^3+z^3\geq 3\sqrt[3]{x^3y^3z^3}=3xyz.$

### Acknowledgment

Dan Sitaru has kindly emailed me a LaTeX file with his solution to the above problem, originally from the School Science and Mathematics Association. The problem is by IULIANA TRAŞCĂ, SCORNICESTI, ROMANIA.