Minimum value of $\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$ If $a,b,c \in R$ find the minimum value of $$E=\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$$
My attempt:
Let $x=a-b,y=b-c,z=c-a$
So $x+y+z=0$
We need to minimize $$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{z}\right)^{4}+\left(\frac{z}{x}\right)^{4}$$
$\implies$
$$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{x+y}\right)^{4}+\left(\frac{x+y}{x}\right)^4$$
Letting $\frac{x}{y}=t$ we get:
$$E=t^4+\frac{1}{(t+1)^4}+\frac{(t+1)^4}{t^4}$$
Now i tried using derivatives, but very tedious.

Hint
Define
$$
x={a-b\over b-c}\quad,\quad y={b-c\over c-a}\quad,\quad z={c-a\over a-b}
$$
and try to minimize $x^4+y^4+z^4$ constrained to $xyz=1$.
Alternatively, you can also define
$$
x=\ln\left|{a-b\over b-c}\right|\quad,\quad y=\ln\left|{b-c\over c-a}\right|\quad,\quad z=\ln\left|{c-a\over a-b}\right|
$$
and minimize $e^{4x}+e^{4y}+e^{4z}$ constrained to $x+y+z=0$.
Edit
Thanks to @RiverLi's comment below, also three more constraints should be included:
$$
{y(x+1)+1=0
\\
z(y+1)+1=0
\\
x(z+1)+1=0
}
$$
