Prove that $\frac{1}{1^4}+\frac{1}{2^4}+\cdots+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ I have to show: $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ for natural $n$
I tried to show it by induction (but I think it could be possible to show it using some ineqaulity of means) so for $n=1$ we have $1=1$ so inequality holds then I assume it's true for $n$ and for $n+1$ my thesis is $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{(n+1)^4} \le 2-\frac{1}{\sqrt{n+1}}$
I know that: 
$\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{(n+1)^4} \le 2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}$ but later I'm not sure if I have to show 
$\displaystyle2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}\le2-\frac{1}{\sqrt{n+1}}$ or should be $\ge$

If we are allowed to use tools from the calculus, we can show that the result holds at $n=1,2$, and then use the fact that
$$\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\cdots\lt \int_2^\infty x^{-4}\,dx.$$
From this we can conclude that our sum $S_n$ is always less than $1+\frac{1}{16}+\frac{1}{24}$.
