Inequality involving logarithms

Question

Prove (or disprove) that $\displaystyle\log_{x}y^z+\log_{y}z^x+\log_{z}x^y>3\;\;$ for $x,y,z>1$.

(This question was inspired by Proving a logarithmic inequality)

Answer 1

$$\log_x\left(y^z\right)=z\cdot\log_xz$$ and $$\log_xz=\dfrac{\log z}{\log x}$$

Now $$z\cdot\dfrac{\log z}{\log x}>0$$ as $x,y,z>1$

Using AM-GM inequality

$$\dfrac{\sum z\dfrac{\log y}{\log x}}3\ge\sqrt[3]{xyz\prod\dfrac{\log y}{\log x}}=\sqrt[3]{xyz}>1$$ as $x,y,z>1$