If $S = 0$, then $Q = 0$, and the inequality is evident. Suppose now that $S$ is not identically zero. Then $\forall x > 0$ $S(x) > 0$. If for some $y > 0$ $P(y) < 0$, then the polynomial $P$, and so the polynomial $S$, have roots on the interval $(0, y)$, which is impossible. So, $P$ and $Q$ are positive for $x > 0$. Rewrite our inequality in the following way:
$$4(2P(x^2)2Q(x^2)) - (2P(x)2Q(x))^2 \le (2P(x^3))^2 + 2(2Q(x^3)).$$
Denote $\alpha = 2P$, $\beta = 2Q$, $\gamma = \alpha\beta = 4PQ$. Then the last inequality becomes:
$$4\gamma(x^2) \le \gamma^2(x) + \alpha^2(x^3) + 2\beta(x^3).$$
Estimate both sides of this inequality:
$$\begin{aligned} \gamma^2(x) + \alpha^2(x^3) + 2\beta(x^3) &= \gamma^2(x) + \beta(x^3) + \alpha^2(x^3) + \beta(x^3) \ge \\ &\ge 4\sqrt[4]{\gamma^2(x)\alpha^2(x^3)\beta^2(x^3)} = 4\sqrt[4]{\gamma^2(x^3)\gamma^2(x)} = 4\sqrt{\gamma(x)\gamma(x^3)}. \end{aligned}$$
If $\gamma(x) = a_0 + a_1x + \dots + a_nx^n$, then
$$(a_0 + a_1x + \dots + a_nx^n)(a_0 + a_1x^3 + \dots + a_nx^{3n}) \ge (\sqrt{a_0a_1} + \sqrt{a_1a_2} + \dots + \sqrt{a_n a_{n+1}})^2$$
(the Cauchy-Schwartz inequality), which implies the required inequality.