This one is simple. Note that, using $n\sqrt{n^2+1}+m\leq 1+\sqrt{2}$, we have, $$ |n\sqrt{n^2+1}-m|=\frac{|n^2(n^2+1)-m^2|}{n\sqrt{n^2+1}+m}\geq \frac{|n^2(n^2+1)-m^2|}{\sqrt{2}+1}. $$Hence, $$ \frac{|n^2(n^2+1)-m^2|}{\sqrt{2}+1}\geq \sqrt{2}-1 \iff |n^2(n^2+1)-m^2|\geq 1. $$Thus, it suffices to make sure, $|n^2(n^2+1)-m^2|\neq 0$. But this is obvious, since had $m^2=n^2(n^2+1)$, we would have $n^2+1$ is a perfect square , contradiction.

First we show that \[|n\sqrt{n^2+1}-m|\geq|n\sqrt{n^2+1}-n^2|.\]Note that \[n^4<n^2(n^2+1)<(n^2+1)^2\]so $\lfloor n\sqrt{n^2+1}\rfloor=n^2$. Thus |n\sqrt{n^2+1}-m| for a fixed $n$ is minimized at either $m=n^2$ or $m=n^2+1$. But observe that \[2n^2+1=\sqrt{4n^4+4n^2+1}>\sqrt{4n^4+4n^2}=2n\sqrt{n^2+1}\]so \[n^2+1-n\sqrt{n^2+1}>n\sqrt{n^2+1}-n^2\]and thus $|n\sqrt{n^2+1}-m|$ is minimized when $m=n^2$. Now we show that \[|n\sqrt{n^2+1}-n^2|\geq\sqrt{2}-1.\]Note that $n$ is increasing and $\sqrt{n^2+1}-n$ is increasing so $n\sqrt{n^2+1}-n^2$ is also increasing. Thus $n\sqrt{n^2+1}-n$ is minimized at $n=1$ which gives the desired inequality. So \[|n\sqrt{n^2+1}-m|\geq|n\sqrt{n^2+1}-n^2|\geq\sqrt{2}-1\]as desired.

oskarbohme34 wrote: Note that $\sqrt{n^2+1}-n$ is increasing This is false... However, one could show that $n\sqrt{n^2+1}-n^2$ is increasing by noting that this approaches $\frac{1}{2}$ from below and gets closer every time.

benstein wrote: oskarbohme34 wrote: Note that $\sqrt{n^2+1}-n$ is increasing This is false... However, one could show that $n\sqrt{n^2+1}-n^2$ is increasing by noting that this approaches $\frac{1}{2}$ from below and gets closer every time. Just on this: an alternative way of seeing that $n\sqrt{n^2+1}-n^2$ is increasing is, $$ n\sqrt{n^2+1}-n^2 = n\frac{1}{\sqrt{n^2+1}+n} = \frac{1}{1+\sqrt{1/n^2+1}}, $$from here, it is clear that, as $n \uparrow$, $n\sqrt{n^2+1}-n^2 \uparrow$.

grupyorum wrote: This one is simple. Note that, using $n\sqrt{n^2+1}+m\leq 1+\sqrt{2}$, why???????

purplish wrote: First we show that \[|n\sqrt{n^2+1}-m|\geq|n\sqrt{n^2+1}-n^2|.\]Note that \[n^4<n^2(n^2+1)<(n^2+1)^2\]so $\lfloor n\sqrt{n^2+1}\rfloor=n^2$. Thus |n\sqrt{n^2+1}-m| for a fixed $n$ is minimized at either $m=n^2$ or $m=n^2+1$. But observe that \[2n^2+1=\sqrt{4n^4+4n^2+1}>\sqrt{4n^4+4n^2}=2n\sqrt{n^2+1}\]so \[n^2+1-n\sqrt{n^2+1}>n\sqrt{n^2+1}-n^2\]and thus $|n\sqrt{n^2+1}-m|$ is minimized when $m=n^2$. Now we show that \[|n\sqrt{n^2+1}-n^2|\geq\sqrt{2}-1.\]Note that $n$ is increasing and $\sqrt{n^2+1}-n$ is increasing so $n\sqrt{n^2+1}-n^2$ is also increasing. Thus $n\sqrt{n^2+1}-n$ is minimized at $n=1$ which gives the desired inequality. So \[|n\sqrt{n^2+1}-m|\geq|n\sqrt{n^2+1}-n^2|\geq\sqrt{2}-1\]as desired. What if $n \sqrt{n^{2}+1}=n^2+0.6$ then $m=n^{2}+1$ minimizes

lian_the_noob12 wrote: What if $n \sqrt{n^{2}+1}=n^2+0.6$ then $m=n^{2}+1$ minimizes Easy to prove that $n^2<n\sqrt{n^2+1}<n^2+\frac{1}{2}$

RagvaloD wrote: lian_the_noob12 wrote: What if $n \sqrt{n^{2}+1}=n^2+0.6$ then $m=n^{2}+1$ minimizes Easy to prove that $n^2<n\sqrt{n^2+1}<n^2+\frac{1}{2}$ How to prove? I am a noob Please Tell me

Just square, to get $n^4<n^4+n^2<n^4+n^2+\frac{1}{4}$