$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$ Can I ask how to solve this type of equation:

$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$$

It is given that $x,y,z>1$. Which properties of the logarithm have to be used?
I know that $\log_a b=\log a/\log b\to \log_{yz}((x^2+4)/(4\sqrt{yz}))=\log ((x^2+4)/(4\sqrt{yz}))/\log xy$
and $\log_a b/c=\log_a b/\log_a c\to \log_{yz}((x^2+4)/(4\sqrt{yz}))=\log_{yz} (x^2+4)/\log_{yz} (4\sqrt{yz})$
And how to solve this type of equation in general?

By AM-GM
$$0=\sum_{cyc}\log_{yz}\frac{x^2+4}{4\sqrt{yz}}\geq\sum_{cyc}\log_{yz}\frac{2\sqrt{x^2\cdot4}}{4\sqrt{yz}}=\sum_{cyc}\left(\log_{yz}x-\frac{1}{2}\right)=$$
$$=\sum_{cyc}\frac{2\ln{x}-\ln{y}-\ln{z}}{2(\ln{y}+\ln{z})}=\frac{1}{2}\sum_{cyc}\frac{\ln{x}-\ln{y}-(\ln{z}-\ln{x})}{\ln{y}+\ln{z}}=$$
$$=\frac{1}{2}\sum_{cyc}(\ln{x}-\ln{y})\left(\frac{1}{\ln{y}+\ln{z}}-\frac{1}{\ln{z}+\ln{x}}\right)=$$
$$=\sum_{cyc}\frac{(\ln{x}-\ln{y})^2}{2(\ln{y}+\ln{z})(\ln{x}+\ln{z})}\geq0.$$
The equality occurs only for $x=y=z=2$ and $\ln{x}=\ln{y}=\ln{z},$ 
which gives that $\{(2,2,2)\}$ is an answer.
