We have
$$(a)\ (6, 0, 0, 0) > (2, 2, 1, 1)$$
$$(b)\ (2, 2, 2, 0) > (2, 2, 1, 1)$$
$$(c)\ (4, 2, 0, 0) > (2, 2, 1, 1)$$
By Muirhead and the above majorizations we have the following
$$1.\ 6(a^6 + b^6 + c^6 + d^6) \geq 4(a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a b^2 c d^2)$$
$$2.\ 6(a^2 b^2 c^2 + a^2 b^2 d^2 + a^2 c^2 d^2 + b^2 c^2 d^2) \geq 4(a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a b^2 c d^2)$$
$$3\ (a^4 d^2 + a^2 b^4 + a^2 d^4 + b^4 d^2 + b^2 d^4 + a^4 c^2 + a^2 c^4 + b^4 c^2 + b^2 c^4 + c^4 d^2 + c^2 d^4 + b^2 a^4) \geq 4(a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a^2 b c d^2 + a b^2 c d^2)$$
And so we have
$$1' \quad (a^6 + b^6 + c^6 + d^6) \geq \frac{4}{6}(a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a b^2 c d^2)$$
$$2' \quad (a^2 b^2 c^2 + a^2 b^2 d^2 + a^2 c^2 d^2 + b^2 c^2 d^2) \geq \frac{4}{6}(a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a b^2 c d^2)$$
$$3' \quad (a^4 d^2 + a^2 b^4 + a^2 d^4 + b^4 d^2 + b^2 d^4 + a^4 c^2 + a^2 c^4 + b^4 c^2 + b^2 c^4 + c^4 d^2 + c^2 d^4 + b^2 a^4) \geq 2(a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a b^2 c d^2)$$
Taking the linear combination $(1') + 6(2') + 3(3')$ we have it that l.h.s. of the resulting inequality becomes $(a^2 + b^2 + c^2 + d^2)^3$ and thus
$$(a^2 + b^2 + c^2 + d^2)^3 \geq \frac{32}{3}(a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a b^2 c d^2).$$
As $a^2 + b^2 + c^2 + d^2 = 1$ the result follows.

Solution 2:

Writing the expression $a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a b^2 c d^2$ as $a b c d (a b + b c + c d + a c + a d + b d)$ and given the condition $a^2 + b^2 + c^2 + d^2 = 1$ we note that $a b c d$ can not exceed $1/16$. Furthermore, using non-negativity of square of $(a-b), (b-c), (c-d), (d-a), (c-a), (d-b)$, and the condition $a^2 + b^2 + c^2 + d^2 = 1$ yields that $a b + b c + c d + a c + a d + b d$ can not exceed $3/2$, and result follows.

Solution 3:

As above, we have
$$a^2 b^2 c d + a b^2 c^2 d + a b c^2 d^2 + a^2 b c^2 d + a^2 b c d^2 + a b^2 c d^2 = a b c d (a b + b c + c d + a c + a d + b d)$$
and $a b c d \leq \frac{1}{16}$ (by AM-GM and $a^2 + b^2 + c^2 + d^2 = 1$). Now
$$a b + b c + c d + a c + a d + b d \leq \frac{a^2+b^2}{2} + \frac{b^2+c^2}{2} + \frac{c^2+d^2}{2} + \frac{a^2+c^2}{2} + \frac{a^2+d^2}{2} + \frac{b^2+d^2}{2}$$
(apply AM-GM to each term of the left hand side). Now the right hand side of the last inequality is equal to $\frac{3}{2}(a^2 + b^2 + c^2 + d^2)$ which, by assumption is $\frac{3}{2}$. The required inequality follows immediately from these observations.