Show that $(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6$ for $a^2 + b^2 + c^2 + d^2 = 1$. For $a, b, c, d \in \Bbb R$ such that $a^2 + b^2 + c^2 + d^2 = 1$, show that $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6.$$
The answer uses the mysterious identity $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 + (a - b)^4 + (a - c)^4 + (a - d)^4 + (b - c)^4 + (b - d)^4 + (c - d)^4 = 6(a^2 + b^2 + c^2 + d^2)^2.$$
But this just comes out of the blue. Are there any other solutions?
Edit (for those who say this is lacking context): This comes from the beginning of a book named Algebraic Inequalities (not in English). After the preface to the first chapter, where the author introduced that many inequalities can be proved by using the non-negativity of $x^2$, this is the first example (without developing any other methods yet). Then the author notes: the reader is supposed to memorize identities like that.

Note that $(a+b)^4+(a-b)^4=2a^4+12a^2b^2+2b^2$. Hence
$\begin{align}(a + b)^4 &+ (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4\\
 &+ (a - b)^4 + (a - c)^4 + (a - d)^4 + (b - c)^4 + (b - d)^4 + (c - d)^4\\
&=6(a^4+b^4+c^4+d^4)+12(a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2)\\
&= 6(a^2 + b^2 + c^2 + d^2)^2.\end{align}$
