Inegalitate rationala.

[Marius Mainea]

Demonstrati inegalitatea:

\( \frac{1\cdot 2\cdot 3}{2^6-1}+\frac{2\cdot 3\cdot 4}{3^6-1}+...+\frac{(n-1)\cdot n\cdot (n+1)}{n^6-1}<\frac{1}{6} \)

O.L.M.2003,Dambovita

[Marius Mainea]

Indicatie :

\( \frac{(k-1)k(k+1)}{k^6-1}=\frac{(k-1)k(k+1)}{(k^3-1)(k^3+1)}=\frac{k}{(k^2+k+1)(k^2-k+1)}=\frac{1}{2}(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}) \)

[Andi Brojbeanu]

\( \frac{1\cdot 2\cdot 3}{2^6-1}+\frac{2\cdot 3\cdot 4}{3^6-1}+...+\frac{(n-1)\cdot n\cdot (n+1)}{n^6-1}=\frac{1}{2}(\frac{1}{2^2-2+1}-\frac{1}{2^2+2+1}+\frac{1}{3^2-3+1}-\frac{1}{3^2+3+1}+....+\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}) \).
\( \frac{1}{k^2+k+1}=\frac{1}{(k+1)^2-(k+1)+1}\Leftrightarrow k^2+k+1=(k+1)(k+1-1)+1\Leftrightarrow k^2+k=k(k+1) \), adevarat.
Atunci membrul stang al ingalitatii devine \( \frac{1}{2}(\frac{1}{4-2+1}-\frac{1}{n^2+n+1})=\frac{1}{2}(\frac{1}{3}-\frac{1}{n^2+n+1}) \).
\( LHS< RHS \Leftrightarrow \frac{1}{3}-\frac{1}{n^2+n+1}< \frac{1}{3}\Leftrightarrow \frac{1}{n^2+n+1}>0 \), adevarat pentru orice \( n\in \mathb{N} \).