Pierādīt, ka
\(\frac{1}{1^{4}+1^{2}+1}+\frac{2}{2^{4}+2^{2}+1}+\frac{3}{3^{4}+3^{2}+1}+\ldots+\frac{2009}{2009^{4}+2009^{2}+1}<\frac{1}{2}\).
Ievērosim, ka katram \(n>0\) pastāv vienādība
\[\begin{aligned} & \frac{n}{n^{4}+n^{2}+1}=\frac{n}{n^{4}+2n^{2}+1-n^{2}}=\frac{n}{\left(n^{2}+1\right)^{2}-n^{2}}=\frac{n}{\left(n^{2}-n+1\right)\left(n^{2}+n+1\right)}= \\ & =\frac{1}{2}\left[\frac{1}{n^{2}-n+1}-\frac{1}{n^{2}+n+1}\right]=\frac{1}{2}\left[\frac{1}{n^{2}-n+1}-\frac{1}{(n+1)^{2}-(n+1)+1}\right] \end{aligned}\]
Saskaitot šīs vienādības pie \(n=1;\ 2;\ 3;\ \ldots;\ 2009\), iegūstam, ka novērtējamās summas vērtība ir \(\frac{1}{2}\left[\frac{1}{1^{2}-1+1}-\frac{1}{2010^{2}-2010+1}\right]<\frac{1}{2} \cdot \frac{1}{1^{2}-1+1}=\frac{1}{2}\), k.b.j.