Update
A possible way is as follows. We use the Buffalo Way. WLOG, assume that $d = \min(a,b,c,d)$.
Let $c=d+s, \ b = d+t, \ a = d+r; \ s, t, r \ge 0$.
We have $\mathrm{LHS} - \mathrm{RHS} = a_8d^8 + a_7d^7 + \cdots + a_1d + a_0$.
To prove (or disprove) that $a_i\ge 0$ for $i=0, 1, 2, \cdots, 8$.
I verified all $a_i$'s by SOS (Sum of Squares) numerical software.
Previously written
This is not an answer. I want to discuss the relationship between two inequalities.
Refer to this link How to prove $\frac{1}{4}(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a})\ge \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}$
Let $a, b, c, d > 0$. Prove that
$$\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^2\ge 4(a^2+b^2+c^2+d^2)+\dfrac{8}{3}[(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2], \qquad (1)$$
$$\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \ge 4\sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}.\qquad\qquad\qquad (2)$$
(1) is stronger than (2) since
\begin{align}
&\Big(4(a^2+b^2+c^2+d^2)+\dfrac{8}{3}[(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2]\Big)^2\\
&\qquad- 64(a^4+b^4+c^4+d^4)\\
=\ & \frac{16}{9} z^TQz\\
\ge\ & 0
\end{align}
where
$$z = \left(\begin{array}{c} { {a}}^2\\ {a} {b}\\ { {b}}^2\\ {a} {c}\\ {b} {c}\\ { {c}}^2\\ {a} {d}\\ {b} {d}\\ {c} {d}\\ { {d}}^2 \end{array}\right), \quad Q = \left(\begin{array}{cccccccccc} 45 & -36 & -4 & -36 & 25 & -4 & -36 & 25 & 25 & -4\\ -36 & 186 & -36 & -45 & -45 & 25 & -45 & -45 & 16 & 25\\ -4 & -36 & 45 & 25 & -36 & -4 & 25 & -36 & 25 & -4\\ -36 & -45 & 25 & 186 & -45 & -36 & -45 & 16 & -45 & 25\\ 25 & -45 & -36 & -45 & 186 & -36 & 16 & -45 & -45 & 25\\ -4 & 25 & -4 & -36 & -36 & 45 & 25 & 25 & -36 & -4\\ -36 & -45 & 25 & -45 & 16 & 25 & 186 & -45 & -45 & -36\\ 25 & -45 & -36 & 16 & -45 & 25 & -45 & 186 & -45 & -36\\ 25 & 16 & 25 & -45 & -45 & -36 & -45 & -45 & 186 & -36\\ -4 & 25 & -4 & 25 & 25 & -4 & -36 & -36 & -36 & 45 \end{array}\right).$$
