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

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.

Answer 1

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: $$ {y(x+1)+1=0 \\ z(y+1)+1=0 \\ x(z+1)+1=0 } $$

Answer 2

We assume $a\ne b\ne c$. We see that if $$E(t)=t^4+\frac{1}{(t+1)^4}+\frac{(t+1)}{t^4},$$ then $$E'(t)=4\frac{(t^2+t+1)^2}{t^5(t+1)^5}p(t),$$ where $$p(t)=t^9+3t^8+t^7-3t^6+t^5+2t^4-10t^3-13t^2-6t-1.$$ From our assumption that $a\ne b\ne c$, we have that $t=\frac{a-b}{b-c}\not\in\{0,-1\}$, and since $a,b,c\in\Bbb R$, $t^2+t+1\ne0$. Thus the critical points of $E(t)$ are the real roots of $p(t)$. There are only three real roots of $p(t)$. Call them $\alpha,\beta,\gamma$, with $\alpha<\beta<\gamma$. We have $$\begin{align} \alpha&\approx -1.7212312143130658672\\ \beta&\approx -0.41902052920932262495\\ \gamma&\approx 1.3865179156901112246. \end{align}$$ Thus $$\begin{align} E(\alpha)&\approx 12.5037895937134016209875697987512072072\color{red}{22979907033168192287594834}\\ E(\beta)&\approx 12.5037895937134016209875697987512072072\color{green}{14601727355467488179851200}\\ E(\gamma)&\approx 12.5037895937134016209875697987512072072\color{blue}{10427320909272240776493390}. \end{align}$$ Thus it appears as if $E(\alpha)\ge E(\beta)\ge E(\gamma)>12+\frac{10803682224985055382}{21444830063582275811}$, so that $$\left(\frac{a-b}{b-c}\right)^4+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^4>12+\frac{10803682224985055382}{21444830063582275811}.$$

Answer 3

The minimum of $E = E(t) = t^4 + \frac{1}{(t+1)^4} + \frac{(t+1)^4}{t^4}, \ t \ne 0, -1$ admits a closed form: $$E_{\min} = \frac{1}{6}\sqrt[3]{98972 + 3396\sqrt{849}} + \frac{1}{6}\sqrt[3]{98972 - 3396\sqrt{849}} + \frac{7}{3}.$$ Note: $E_{\min} \approx 12.5037895937134016209875697987512$.

Proof: Denote $$Q = \frac{1}{6}\sqrt[3]{98972 + 3396\sqrt{849}} + \frac{1}{6}\sqrt[3]{98972 - 3396\sqrt{849}} + \frac{7}{3}.$$ We have $$Q^3-7Q^2+3Q-898 = 0, \quad Q-7 > 0, \quad Q^2 - 7Q + 3 > 0. \tag{1}$$

We have \begin{align} &[E^2 + (Q-7)E + Q^2-7Q+3](E - Q)\\ =\ & E^3 - 7E^2 + 3E - Q^3 + 7Q^2 - 3Q\\ =\ & E^3 - 7E^2 + 3E - 898\\ =\ & \left(t^4 + \frac{1}{(t+1)^4} + \frac{(t+1)^4}{t^4}\right)^3 - 7\left(t^4 + \frac{1}{(t+1)^4} + \frac{(t+1)^4}{t^4}\right)^2\\ &\qquad + 3\left(t^4 + \frac{1}{(t+1)^4} + \frac{(t+1)^4}{t^4}\right) - 898\\ =\ & \frac{1}{(t+1)^{12}t^{12}}f(t)[g(t)]^2\\ \ge\ & 0 \tag{2} \end{align} where \begin{align} f(t) &= t^{18} + 6 t^{17} + 19 t^{16} + 40 t^{15} + 57 t^{14} + 60 t^{13} + 93 t^{12} + 308 t^{11}\\ &\quad + 901 t^{10} + 1832 t^9 + 2650 t^8 + 2836 t^7 + 2316 t^6 + 1466 t^5\\ &\quad + 717 t^4 + 264 t^3 + 70 t^2 + 12 t + 1, \\ g(t) &= t^9 + 3t^8 + t^7 - 3t^6 + t^5 + 2t^4 - 10t^3 - 13t^2 - 6t - 1, \end{align} and it is not difficult to prove that $f(t) > 0$ for all $t$ in $\mathbb{R}$.

Since $E^2 + (Q-7)E + Q^2-7Q+3 > 0$, from (2), we have $E \ge Q$. Also, $g(t) = 0$ has at least one real root, and thus the minimum of $E$ is exactly equal to $Q$. We are done.

Answer 4

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: $E_{\min} \approx 12.5037895937134016209875697987512$.