Show that $\pi-\left(90\sum_{n=1}^mn^{-4}\right)^{\frac14}<\pi-\left(6\sum_{n=1}^mn^{-2}\right)^{\frac12}$ I was bored and I found the inequality
$$\pi-\left(90\sum_{n=1}^mn^{-4}\right)^{\frac14}<\pi-\left(6\sum_{n=1}^mn^{-2}\right)^{\frac12},$$
where $m$ is a positive integer.
Which is basically derived from $\zeta(4)=\frac{\pi^4}{90}$ and $\zeta(2)=\frac{\pi^2}6$.
This is equivalent to
$$5\sum_{n=1}^mn^{-4}>2\left(\sum_{n=1}^mn^{-2}\right)^2.$$
At first I tried a direct proof, and simplified it down to
$$3\sum_{n=1}^mn^{-4}>4\sum_{\substack{1\le i,j\le m\\i\neq j}}(ij)^{-2}$$
but there wasn't an obvious way for me to proceed from here.
My second attempt was to use induction, and the case $m=1$ was trivial, but the IH was too weak. I used the hypothesis to get
$$5(m+1)^{-4}+5\sum_{n=1}^mn^{-4}>5(m+1)^{-4}+2\left(\sum_{n=1}^mn^{-2}\right)^2,$$
but $2\left(\sum_{n=1}^mn^{-2}+(m+1)^{-2}\right)^2\ge RHS$ (by desmos), so that didn't work as well. I don't think Cauchy works either, though not that sure.
If anyone could provide a solution, then I'd greatly appreciate it! I preferably want an elementary solution.
 A: $5\sum_{n=1}^mn^{-4}>2\left(\sum_{n=1}^mn^{-2}\right)^2.
$
$5(\frac{\pi^4}{90}-\sum_{n=m+1}^{\infty}n^{-4})>2(\frac{\pi^2}{6}-\left(\sum_{n=m+1}^{\infty}n^{-2})\right)^2).
$
$5(\frac{\pi^4}{90}-t_4(m))>2(\frac{\pi^2}{6}-t_2(m))^2.
$
$\frac{\pi^4}{18}-5t_4(m)
\gt 2(\frac{\pi^4}{36}-2\frac{\pi^2}{6}t_2(m)-t_2^2(m))\\
= \frac{\pi^4}{18}-\frac{2\pi^2}{3}t_2(m)-2t_2^2(m)\\
$
Interesting the $\pi^2/18$
cancels out.
$5t_4(m)
\lt \frac{2\pi^2}{3}t_2(m)+2t_2^2(m)\\
$
According to
Computing the tail of the zeta function $\sum_{n>x}n^{-s}$,
$t_s(n)
=\frac{n^{1-s}}{s-1}\left(1-\frac{s-1}{2n}+O\left(\frac1{n^2}\right)\right).
$
Therefore
$t_2(n)
=\frac{n^{-1}}{1}\left(1-\frac{1}{2n}+O\left(\frac1{n^2}\right)\right)
=\frac1{n}\left(1-\frac{1}{2n}+O\left(\frac1{n^2}\right)\right)
$
and
$t_4(n)
=\frac{n^{-3}}{3}\left(1-\frac{3}{2n}+O\left(\frac1{n^2}\right)\right)
$.
Using the first term of each,
$t_2(m)
\approx \dfrac1{m}
$
and
$t_4(m)
\approx \dfrac1{3n^3}
$.
Putting these in,
ths inequality becomes
$5 \dfrac1{3m^3}
\lt \dfrac{2\pi^2}{3}\dfrac1{m}+2\dfrac1{m^2}
$
or
$\dfrac53
\lt \dfrac{2\pi^2m^2}{3}+2m
$
and this is true for all $m$.
I'll leave it at this.
A: Let
$$f(m) = 5\sum_{n=1}^m n^{-4} - 2\left(\sum_{n=1}^m n^{-2}\right)^2.$$
We have
$$f(m+1) = 5\sum_{n=1}^m n^{-4} + \frac{5}{(m+1)^4} -
2\left(\sum_{n=1}^m n^{-2} + \frac{1}{(m+1)^2}\right)^2 $$
and
\begin{align*}
 f(m) - f(m+1) &= \frac{4}{(m+1)^2}\sum_{n=1}^m n^{-2} - \frac{3}{(m+1)^4}\\
 &\ge \frac{4}{(m+1)^2} m^{-2} - \frac{3}{(m+1)^4}\\
 & > 0.
\end{align*}
Also, $\lim_{m\to \infty} f(m)
= 5\zeta(4) - 2[\zeta(2)]^2 = 
5\cdot \frac{\pi^2}{90} - 2\cdot (\frac{\pi^2}{6})^2 = 0$.
Thus, $f(m) > 0$ for all $m\ge 1$.
A: At leats, for large $m$, it is easy to show $$f(m) = 5\sum_{n=1}^m n^{-4} - 2\left(\sum_{n=1}^m n^{-2}\right)^2=5 H_m^{(4)}-2 \left(H_m^{(2)}\right){}^2$$ Using asymptotics
$$f(m)=\frac{2 \pi ^2}{3 m}-\frac{6+{\pi ^2}}{3m^2}+\frac{3+\pi ^2}{9
   m^3}+O\left(\frac{1}{m^4}\right)$$ which is positive for all $m$.
