$ \frac{xyz}{(x+y)(y+z)(x+z)} \leq \frac{1}{8} $

Question

Show that for positive real numbers $ x $, $ y $, and $ z $, the following inequality holds:

$
\frac{xyz}{(x+y)(y+z)(x+z)} \leq \frac{1}{8}
$

Attempt:

I know that $\frac{x+y}{2} \geq \sqrt{xy}$, similarly for $y$ and $z$. From this, it follows that $\frac{1}{x+y} \leq \frac{1}{2\sqrt{xy}}$ (similarly for $y$ and $z$). From this, it follows:

$
\frac{xyz}{(x+y)(y+z)(x+z)} \leq \frac{xyz}{8xyz} = \frac{1}{8}
$

Is this procedure correct? Is there another (maybe easier) solution?

Answer 1

This is from scratch

$\sqrt[3]{xyz} = \sqrt{\sqrt[3]{x\cdot y}} \cdot \sqrt{\sqrt[3]{y\cdot z}} \cdot \sqrt{\sqrt[3]{z\cdot x}} \overset{\text{AM-GM}}{\leqslant} \frac{\sqrt[3]{x} + \sqrt[3]{y}}{2} \cdot\frac{\sqrt[3]{y} + \sqrt[3]{z}}{2}\cdot\frac{\sqrt[3]{z} + \sqrt[3]{x}}{2}$

$2^3 \sqrt[3]{xyz} \leqslant (\sqrt[3]{x} +\sqrt[3]{y})(\sqrt[3]{y} +\sqrt[3]{z})(\sqrt[3]{z} +\sqrt[3]{x})$

$(x,y,z) = (a^3 , b^3 , c^3 ) $

$\color{blue}{\boxed{8abc \leqslant (a+b)(b+a)( c+a)}} $

Equality when $a=b , b=c , c=a $

$a=b=c$

$8c^3 = (2c)(2c)(2c)$