# Another Inequality with Logarithms, But Not Really

### Solution

Let $a=\log_y x;\,$ $b=\log_z y;\,$ $c=\log_x z.\,$ Since $\log_zx=\log_zy\cdot\log_yx=ab,\,$ the inequality reduces to

$\displaystyle \sum \frac{a^3+b^3}{a^2+ab+b^2}\geq 2,$

with $abc=1.$

To continue,

\displaystyle \begin{align} \frac{a^3+b^3}{a^2+ab+b^2}&=\frac{(a+b)(a^2-ab+b^2)}{a^2+ab+b^2}\\ &\geq \frac{a+b}{3} \end{align}

because $\displaystyle \frac{a^2-ab+b^2}{a^2+ab+b^2}\geq \frac{1}{3},\,$ which is equivalent to $2(a-b)^2\geq 0.$

Thusm using the AM-GM inequality,

\displaystyle \begin{align} \sum \frac{a^3+b^2}{a^2+ab+b^2}&\geq \sum \frac{a+b}{3}=\frac{2}{3}(a+b+c)\\\ &\overbrace{\geq}^{AM-GM} \frac{2}{3}\cdot 3\sqrt[3]{abc}=2. \end{align}

### Acknowledgment

Dan Sitaru has kindly reposted the above problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later emailed me his solution in a LaTeX file.