If some configuration works, say $a - b - c,$ then its opposite also works, which is $-a+b+c,$ since they have the same magnitude. If a configuration doesn't work, its opposite doesn't work (likewise). Thus, we must only consider $a+b+c,-a+b+c,a-b+c,a+b-c.$ Consider the squares of the magnitudes. We must show that one of them is less than or equal to two. The squared magnitude of $a+b+c$ using the dot product and its linearity is $3 + 2(a \bullet b + b \bullet c + c \bullet a).$ The squared magnitude of $-a+b+c$ is $3 + 2(-a \bullet b + b \bullet c - c \bullet a).$ Symmetrically, we have $\| a-b+c \|^2 = 3 + 2(-a \bullet b - b \bullet c + c \bullet a)$ and $\| a+b-c \|^2 = 3 + 2(a \bullet b - b \bullet c - c \bullet a).$ Adding all four up, we get $12,$ and all the dot products cancel nicely. This means at least one of the squared magnitudes is less than or equal to $3.$ Thus, at least one $|x|$ is less than or equal to $\sqrt3.$ Thus, not only is $|x| \le \sqrt2$ always achievable, so is $|x| \le \sqrt3.$