By the inequality between the arithmetic and the geometric mean, we have
$$(a+b)(b+c)(c+d)(d+a) \leq \left( \frac{(a+b)+(b+c)+(c+d)+(d+a)}{4} \right)^4 = 2^4 \left( \frac{a+b+c+d}{4} \right)^4.$$
By the inequality between the quadratic and the arithmetic mean, we have
$$2^4 \left( \frac{a+b+c+d}{4} \right)^4 \leq 2^4 \left( \frac{a^2+b^2+c^2+d^2}{4} \right)^2 = (a^2+b^2+c^2+d^2)^2,$$
as required.
In the second inequality, equality holds if and only if $a = b = c = d$, but in that case, equality holds also in the original inequality. Therefore, equality holds if and only if $a = b = c = d$.