China Mathematical Competition

As $$1+{\log }_2(x^4+1)={\log }_2(2x^4+2),$$ and ${\log }_{2}y$ is monotonically increasing over $(0,+\infty )$, the given inequality is equivalent to $$x^{12}+3x^{10}+5x^8+3x^6+1<2x^4+2$$ or $$x^{12}+3x^{10}+5x^8+3x^6-2x^4-1<0.$$ It can be rewritten as $$\begin{array}{rl}& x^{12}+x^{10}-x^8\\ & +2x^{10}+2x^8-2x^6\\ & +4x^8+4x^6-4x^4\\ & +x^6+x^4-x^2\\ & +x^4+x^2-1<0.\end{array}$$ That is to say, $$(x^8+2x^6+4x^4+x^2+1)(x^4+x^2-1)<0.$$ Then we have $x^4+x^2-1<0$. It follows that $x^2<\frac{-1+\sqrt{5}}{2}$, i.e. $$-\sqrt{\frac{-1+\sqrt{5}}{2}}<x<\sqrt{\frac{-1+\sqrt{5}}{2}}.$$ So the solution set is $(-\sqrt{\frac{-1+\sqrt{5}}{2}},\sqrt{\frac{-1+\sqrt{5}}{2}})$.

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Alternative solution.

As $$1+{\log }_2(x^4+1)={\log }_2(2x^4+2),$$ and ${\log }_{2}y$ is monotonically increasing over $(0,+\infty )$, the given inequality is equivalent to $$x^{12}+3x^{10}+5x^8+3x^6+1<2x^4+2$$ or $$\begin{array}{rl}{\left(\frac{1}{x^2}\right)}^3+2\left(\frac{1}{x^2}\right)& >x^6+3x^4+3x^2+1+2x^2+2\\ & =(x^2+1{)}^3+2(x^2+1).\end{array}$$ Define $g(t)=t^2+2t$. Then we have $$g\left(\frac{1}{x^2}\right)>g(x^2+1).$$ Obviously, $g(t)$ is a monotonically increasing function; then we have $$\frac{1}{x^2}>x^2+1.$$ That is to say, $$x^4+x^2-1<0.$$ We obtain $x^2<\frac{-1+\sqrt{5}}{2}$. So the solution set is $(-\sqrt{\frac{-1+\sqrt{5}}{2}},\sqrt{\frac{-1+\sqrt{5}}{2}})$.