Belarusian Mathematical Olympiad

Let $M(n,m)=|n\sqrt{n^2+1}-m|$. Using an obvious chain of inequalities $$n^2<n\sqrt{n^2+1}<n^2+\frac{1}{2},$$ we obtain $$n^2-m<n\sqrt{n^2+1}-m<n^2-m+\frac{1}{2}.$$ Therefore, if $m\ne n^2$ then $M(n,m)>1/2$. Since $1/2>\sqrt{2}-1$, the required inequality is proved for all $m\ne n^2$.

Let $m=n^2$. Consider the function $f(n)=M(n,n^2)=n\sqrt{n^2+1}-n^2$. We will show that it increases at $n>0$. Indeed, for all $a>b>0$ the difference $f(a)-f(b)$ can be transformed: $$\begin{array}{rl}& (a\sqrt{a^2+1}-a^2)-(b\sqrt{b^2+1}-b^2)=(a\sqrt{a^2+1}-b\sqrt{b^2+1})-(a^2-b^2)=\\ & =\frac{a^2(a^2+1)-b^2(b^2+1)}{a\sqrt{a^2+1}+b\sqrt{b^2+1}}-(a^2-b^2)=(a^2-b^2)\left(\frac{a^2+b^2+1}{a\sqrt{a^2+1}+b\sqrt{b^2+1}}-1\right).\end{array}(1)$$ As mentioned above, $a^2+\frac{1}{2}>a\sqrt{a^2+1}$ and $b^2+\frac{1}{2}>b\sqrt{b^2+1}$. Summing these inequalities and substituting to (1) we obtain that $f(a)-f(b)>0$. Therefore, the minimum of the function $f(n)$ is achieved at $n=1$ and equals $\sqrt{2}-1$.