Note that for all real $x$
$$x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1) = (x^2 + 1)(x^2 + \sqrt{3}x + 1)(x^2 - \sqrt{3}x + 1),$$
is a product of three positive numbers, since $x^2 \pm \sqrt{3}x + 1 = (x \pm \frac{\sqrt{3}}{2})^2 + \frac{1}{4}$.
We use AGM for three positive variables: $\sqrt[3]{ABC} \le \frac{A+B+C}{3}$ with $A = \frac{x^2+1}{2}$, $B = a(x^2 + \sqrt{3}x + 1)$ and $C = b(x^2 - \sqrt{3}x + 1)$, where $a, b$ are positive numbers (to be chosen) such that $ab = 1$. Then
$$\begin{aligned} \sqrt[3]{\frac{x^6 + 1}{2}} &\le \frac{\frac{x^2+1}{2} + a(x^2 + \sqrt{3}x + 1) + b(x^2 - \sqrt{3}x + 1)}{3} \\ &= \frac{(1 + 2(a + b))x^2 + 2\sqrt{3}(a - b)x + 1 + 2(a + b)}{6} \\ &= \frac{3x^2 - 4x + 3}{2}, \end{aligned}$$
if, in addition, $a, b$ satisfy the equations $1 + 2(a + b) = 9$, $\sqrt{3}(a - b) = -6$. The numbers $a = 2 - \sqrt{3}$, $b = 2 + \sqrt{3}$ satisfy these and $ab = 1$. Hence,
$$\sqrt[3]{\frac{x^6 + 1}{2}} \le \frac{3x^2 - 4x + 3}{2},$$
for all real $x$, with equality iff
$$\frac{x^2 + 1}{2} = a(x^2 + \sqrt{3}x + 1) = b(x^2 - \sqrt{3}x + 1),$$
i.e., $x = 1$.