Finding the minimum value of $\log _d a + \log _bd + \log _ac + \log _c b$

Question

The question says to find the minimum value of $$\log _da + \log _bd + \log _ac + \log _cb$$ Given, $a,b,c,d \; \in R^+ -\Bigl(1\Bigl)$
My approach:
I used the AM-GM inequality and so we can see that $$\frac{\log _da + \log _bd + \log _ac + \log _cb}{4} \geq \Biggl(\log _da \cdot \log _bd \cdot \log _ac \cdot \log_cb\Biggl)^\frac{1}{4}$$ So the minimum value is, $$4\Biggl(\log _da \cdot \log _bd \cdot \log _ac \cdot \log _cb\Biggl)^\frac{1}{4}$$
Now, I wrote it as, $$\Biggl(\frac{\log_da}{\log_db} \cdot \frac{\log_ca}{\log_cb}\Biggl)^\frac{1}{4}$$
I am not sure what do to after this point. I also feel like we can use HM inequality but that will only give us,
$$\log _da + \log _bd + \log _ac + \log _cb = \log _ad + \log _db + \log _ca + \log _bc $$

Answer 1

By Rearrangement inequality, since $\log$ is an increasing function, we have

$$\log _da + \log _bd + \log _ac + \log _cb=$$ $$=\frac{\log a}{\log d}+\frac{\log d}{\log b}+\frac{\log c}{\log a}+\frac{\log b}{\log c}\ge \frac{\log a}{\log a}+\frac{\log b}{\log b}+\frac{\log c}{\log c}+\frac{\log d}{\log d}=4$$