TITLE: Minimum value of $\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: My second solution (using derivative):
Since $\lim_{t\to \infty} E(t) = \infty$, $\lim_{t\to - \infty} E(t) = \infty$,
$\lim_{t\to 0} E(t) = \infty$ and $\lim_{t\to -1} E(t) = \infty$, the minimum of $E(t)$
is attained at the points with $E'(t) = 0$.
We have
\begin{align}
E'(t) &= 
\frac{4(t^2+t+1)^2}{(t+1)^5t^5}(t^9+3t^8+t^7-3t^6+t^5+2t^4-10t^3-13t^2-6t-1)\\
&= \frac{4(t^2+t+1)^2}{(t+1)^5t^5}F(t)[t^3+(c+3)t^2+ct-1]
\end{align}
where
$$c = -\frac{1}{6}\sqrt[3]{108+12\sqrt{849}} + \frac{1}{6}\sqrt[3]{-108 + 12\sqrt{849}}-2,$$
and
\begin{align}
F(t) &= t^6-ct^5+(c+1)^2 t^4+(-c^3-4c^2-7c-5)t^3\\
&\quad +(c^2+3c+4)^2t^2+(-c^5-8c^4-31c^3-67c^2-81c-45)t+1.
\end{align}
It is not difficult to prove that $F(t) > 0$ for all $t$ in $\mathbb{R}$.
Thus, all real roots of $E'(t) = 0$ are the real roots of $t^3+(c+3)t^2+ct-1 = 0$.
By using $t^3+(c+3)t^2+ct-1 = 0$, we have
\begin{align}
E(t) &= \frac{t^8(t+1)^4+t^4+(t+1)^8}{(t+1)^4t^4}\\
&= \frac{\mathrm{rem}(t^8(t+1)^4+t^4+(t+1)^8,\ t^3+(c+3)t^2+ct-1)}{\mathrm{rem}((t+1)^4t^4,\ t^3+(c+3)t^2+ct-1)}\\
&= c^4+8c^3+32c^2+68c+69\\
&= \frac{1}{6}\sqrt[3]{98972 + 3396\sqrt{849}}
+ \frac{1}{6}\sqrt[3]{98972 - 3396\sqrt{849}} + \frac{7}{3}
\end{align}
where $\mathrm{rem}(p(x),q(x))$ is the remainder of $p(x)$ divided by $q(x)$ for two polynomials $p(x), q(x)$.
Thus, the minimum of $E = E(t) = t^4 + \frac{1}{(t+1)^4} + \frac{(t+1)^4}{t^4}, \ t \ne 0, -1$ is given by
$$E_{\min} = \frac{1}{6}\sqrt[3]{98972 + 3396\sqrt{849}}
+ \frac{1}{6}\sqrt[3]{98972 - 3396\sqrt{849}} + \frac{7}{3}.$$
Note: $E_{\min} \approx 12.5037895937134016209875697987512$.