Inequality for $|x| \geq 1$

Question

For $|x|\geq 1$, is the following inequality true? $$ |x|^8\leq(1+(|x|^2-1)^4)^\frac{n+2s}{8}. $$ I have tried assuming such an inequality holds and end up checking with the following inequality for $|x|\geq 1$: $$ \frac{2x^4}{x^4+1}\geq 2x^2-1. $$ Still there is doubt whether the second hold. Can somebody please help me solving this issue. Thanks. Here $n\in\mathbb{N}$ and $0<s<1$.

Answer 1

I think that the inequality does not hold.

If $|x| > 1$, the inequality is written as $$\frac{4\ln |x|^2}{\ln (1 + (|x|^2-1)^4)} \le \frac{n+2s}{8}.$$ This inequality does not hold since $$\lim_{|x|^2 \to 1^{+}} \frac{4\ln |x|^2}{\ln (1 + (|x|^2-1)^4)} = \lim_{y\to 1^{+}} \frac{4\ln y}{\ln (1 + (y-1)^4)} = \lim_{y\to 1^{+}} \frac{\frac{4}{y}}{\frac{4(y-1)^3}{1 + (y-1)^4}} = \infty$$ where L'Hopital's Rule is used.