Solution:
The inequality is equivalent to
$$\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)^{2} \geq\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\right)^{2}$$
By the Cauchy-Schwarz inequality,
$$\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\right) \geq\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\right)^{2}$$
Hence it is sufficient to prove that
$$\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)^{2} \geq\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\right)$$
i.e. to prove $a b^{3}+b c^{3}+c d^{3}+d a^{3} \geq a^{3} b+b^{3} c+c^{3} d+d^{3} a$.
This inequality can be written successively
$$a\left(b^{3}-d^{3}\right)+b\left(c^{3}-a^{3}\right)+c\left(d^{3}-b^{3}\right)+d\left(a^{3}-c^{3}\right) \geq 0$$
or
$$(a-c)\left(b^{3}-d^{3}\right)-(b-d)\left(a^{3}-c^{3}\right) \geq 0$$
which comes down to
$$(a-c)(b-d)\left(b^{2}+b d+d^{2}-a^{2}-a c-c^{2}\right) \geq 0$$
The last inequality is true because $a-c \leq 0, b-d \leq 0$, and $\left(b^{2}-a^{2}\right)+(b d-a c)+\left(d^{2}-c^{2}\right) \geq 0$ as a sum of three non-negative numbers.
The last inequality is satisfied with equality whence $a=b$ and $c=d$. Combining this with the equality cases in the Cauchy-Schwarz inequality we obtain the equality cases for the initial inequality: $a=b=c=d$.

Remark. Instead of using the Cauchy-Schwarz inequality, once the inequality $a b^{3}+b c^{3}+c d^{3}+$ $d a^{3} \geq a^{3} b+b^{3} c+c^{3} d+d^{3} a$ is established, we have $2\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right) \geq\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)+\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\right)=\left(a b^{3}+a^{3} b\right)+\left(b c^{3}+b^{3} c\right)+\left(c d^{3}+c^{3} d\right)+\left(d a^{3}+d^{3} a\right) \stackrel{A M-G M}{\geq}$ $2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} d^{2}+2 d^{2} a^{2}$ which gives the conclusion.