Log Inequality

We expand $\log_y(xz)\cdot \log_z(xy)$ to get $(\log_yx + \log_yz)(\log_z x + \log_zy).$ We expand to get $\log_yx\cdot\log_zx + \log yz \cdot \log_zx + \log zy \cdot \log yx + \log yz \cdot \log zy$ which becomes $(\log_y x)(\log_z x) + (\log_y x) + (\log_z x) + 1$ which factors as $(\log_y x+ 1)(\log_zx + 1)$ Make $y$ and $x$ as large as possible and the logs approach 0. Thus our minimal value is 1.

We want to find the minimum of $\log_y(xz)\cdot\log_z(xy)$ .

First, we expand the logs: $\log_y(xz)\cdot\log_z(xy)=(\log_yx+\log_yz)(\log_zx+\log_zy)$

Note that $\log_yz=\dfrac{1}{\log_zy}$ ; and the less variables we have, the better. So let's substitute: $(\log_yx+\log_yz)(\log_zx+\log_zy)=\left(\log_yx+\dfrac{1}{\log_zy}\right)(\log_zx+\log_zy)$

Note that we already have a $\log_yx$ and a $\log_zy$ ; all we need to do is have a $\log_xz$ to perform a clever substitution. Fortunately, this is easy: $\left(\log_yx+\dfrac{1}{\log_zy}\right)(\log_zx+\log_zy)=\left(\log_yx+\dfrac{1}{\log_zy}\right)\left(\log_zy+\dfrac{1}{\log_xz}\right)$

Now we substitute $a=\log_yx$ , $b=\log_zy$ , and $c=\log_xz$ . Note that this brings on the condition that $abc=1$ . Also, since $x,y,z > 1$ , we have $a,b,c > 0$ .

Thus, we want to find the minimum of $\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)$

Expanding: $\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)=ab+\dfrac{a}{c}+\dfrac{1}{bc}+1$

Using $abc=1$ : $ab+\dfrac{a}{c}+\dfrac{1}{bc}+1=a+\dfrac{1}{c}+\dfrac{a}{c}+1$

Now it is obvious that the minimum of this can be achieved for $a\to 0$ and $c\to \infty$ . When this happens, the minimum is $1$ , so our final answer is $\boxed{1.000}$ . Note that this minimum cannot be achieved.

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A possible equality case is $(a,b,c)=(2^{-n},1,2^n)$ as $n\to \infty$

This corresponds to $x^{2^n}=y=z$ . We arbitrarily let $x=2$ , so a possible equality case for $x,y,z$ is $\boxed{(x,y,z)=\left(2,2^{2^n},2^{2^n}\right)}$ as $n\to \infty$ .

In fact, any setup where $x \ll y\le z$ works. (WLOG $y\le z$ )