Problems of Ukrainian Authors

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: $$\begin{array}{rl}{\gamma }^2(x)+{\alpha }^2(x^3)+2\beta (x^3)& ={\gamma }^2(x)+\beta (x^3)+{\alpha }^2(x^3)+\beta (x^3)\ge \\ & \ge 4\sqrt[4]{{\gamma }^2(x){\alpha }^2(x^3){\beta }^2(x^3)}=4\sqrt[4]{{\gamma }^2(x^3){\gamma }^2(x)}=4\sqrt{\gamma (x)\gamma (x^3)}.\end{array}$$ If $\gamma (x)=a_0+a_{1}x+\cdots +a_n{x}^n$, then $$(a_0+a_{1}x+\cdots +a_n{x}^n)(a_0+a_1{x}^3+\cdots +a_n{x}^{3n})\ge (\sqrt{a_0{a}_1}+\sqrt{a_1{a}_2}+\cdots +\sqrt{a_n{a}_{n+1}}{)}^2$$ (the Cauchy-Schwartz inequality), which implies the required inequality.