Prove $\frac{((1+(|k|)^{1/8})^{4n}}{n(n-1)(2n-1)(2n-3)}\leq \frac{(4n-1)(4n-3)(4n-5)(4n-7)}{630}|k|$

Question

Determine whether it is true or false.

There exists some $k\in \Bbb R,n\in \Bbb N$ \ {$0,1$} such that $\frac{(1+(|k|)^{1/8})^{4n}}{n(n-1)(2n-1)(2n-3)}\leq \frac{(4n-1)(4n-3)(4n-5)(4n-7)}{630}|k|$.

I tried to expand the term $(1+(|k|)^{1/8})^{4n}$, and get like $1+4n(\sqrt[8]{|k|})+\frac{4n(4n-1)}{2}(\sqrt[8]{|k|})^2+...+4n(\sqrt[8]{|a|})^{4n-1}+(\sqrt[8]{|a|})^{4n}$.
How should I continue the proof? And how can I get other terms (e.g. the terms in RHS) so as to compare them?
I mean how to get the 'inequality' in the statement?
Thanks!

P.S: Is Bernoulli's inequality useful? I only know $(1+\sqrt[8]{|a|})^{4n}\geq1+4n\sqrt[8]{|a|}.$

Answer 1

At $k=0$, the inequality doesn't hold, as the left side is positive and the right side is nonzero. Now, for $k\neq 0$, the inequality is the same as $$\frac{(1+x)^{4n}}{x^8}\leq \frac{n(n-1)(2n-1)(2n-3)(4n-1)(4n-3)(4n-5)(4n-7)}{630}$$ for $x>0$ under the substitution $x=|k|^{1/8}$ and some reordering of terms. However, $$(1+x)^{4n}>x^8\binom{4n}8$$ by the binomial theorem, and \begin{align*} \binom{4n}{8}&=\frac{(4n)(4n-1)(4n-2)(4n-3)(4n-4)(4n-5)(4n-6)(4n-7)}{40320}\\ &=\frac{n(n-1)(2n-1)(2n-3)(4n-1)(4n-3)(4n-5)(4n-7)}{630}. \end{align*} So, the desired inequality never holds.