$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$ 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?
 A: Notice that we can use the following property $\log_a \frac{b}{c}=\log_a b-\log_a c$ to get:
$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)=\log_{yz}\frac{x^2+4}{4}-\log_{yz}\sqrt{yz}=\log_{yz}\frac{x^2+4}{4}-\frac{1}{2}$$
The equation can, thus, be written as:
$$\log_{yz}\frac{x^2+4}{4}+\log_{zx}\frac{y^2+4}{4}+\log_{xy}\frac{z^2+4}{4}=\frac{3}{2}$$
Notice that for any real number $a$, we have:
$$\frac{a^2+4}{4}\geq a \Leftrightarrow (a-2)^2\geq 0$$
with equality only if $a=2$. Therefore:
$$\log_{yz}\left(\frac{x^2+4}{4}\right)\ge \log_{yz} x=\frac{\ln x}{\ln y+\ln z}$$
Summing up, we arrive at:
$$\frac{3}{2}\geq \frac{\ln x}{\ln y+\ln z}+\frac{\ln y}{\ln x+\ln z}+\frac{\ln z}{\ln x+\ln y}$$
If we let $a=\ln x, b=\ln y, c=\ln z$, since $x,y,z>1$, we have $a,b,c>0$ and:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq \frac{3}{2}$$
However, Nesbitt's inequality states that for any positive real numbers $a,b,c$, we must have:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}$$
with equality only if $a=b=c$. This implies immediately that the only solution of the equation is $x=y=z=2$.
Note: As far as I know, there is no standard way to solve this sort of multivariable logarithm equations. 
Most of the time, you have to use inequalities and show that the equation is a particular equality case. In fact, it given that $x,y,z>1$ so that the logarithm will be positive ($\log_a b$ is positive if $a,b>1$ or $a,b<1$).
A: Hint : First use:  $a^2 + 4 \ge 4a$ for each term on the left. Then split the log and show next that the left is $\ge 0$. Equality is at $ x = y = z = 2$.
A: 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.
