Inequality

Let $x, y,z>0$ such that $x+y+z+2=xyz$ . Prove that:

$$x+y+z+6 \geq 2\left( \sqrt{xy}+ \sqrt{yz}+\sqrt{zx}\right)$$

Solution

Let

$$a=\frac{1}{1+x} \quad, \quad b=\frac{1}{1+y} \quad, \quad c=\frac{1}{1+z}$$

Using $x+y+z+2=xyz$ we get that $a+b+c=1$. Thus, $x=\frac{1-a}{a}=\frac{b+c}{a}$ and similarly $y=\frac{c+a}{b}, \; z=\frac{a+b}{c}$. Applying the AM - GM inequality we have that:

\begin{align*}
x+y+z+6 &=\sum \frac{b+c}{a}+6 \\
 &= \sum \left ( \frac{c+a}{c}+ \frac{a+b}{b} \right )\\
 &\geq 2 \sum \sqrt{\frac{\left ( c+a \right )\left ( a+b \right )}{bc}} \\
 &= 2\left ( \sqrt{xy}+ \sqrt{yz}+ \sqrt{zx} \right )
\end{align*}

This exercise was taken from the journal Excalibur.