Show that $\pi-\left(90\sum_{n=1}^mn^{-4}\right)^{\frac14}<\pi-\left(6\sum_{n=1}^mn^{-2}\right)^{\frac12}$

$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.

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 $$f(m) - f(m+1) = \frac{4}{(m+1)^2}\sum_{n=1}^m n^{-2} - \frac{3}{(m+1)^4} > 0.$$ Also, $\lim_{m\to \infty} f(m) = 0$. Thus, $f(m) > 0$ for all $m\ge 1$.