Inegalitate interesanta (proprie)

[BogdanCNFB]

Fie \( n\in N^* \) si \( a_k\ge n^{2k},k=\overline{1,n} \). Demonstrati ca:
\( 2n^n\cdot\sum_{k=1}^n\sqrt{a_k-n^{2k}}\leq\sum_{k=1}^na_k\cdot n^{n-k} \).

[Marius Mainea]

Notam \( \sqrt{a_k-n^{2k}}=b_k \) si inegalitatea devine \( 2\sum b_k\leq \sum \frac{b_k^2+n^{2k}}{n^k} \) care se demonstreza de exemplu folosind inegalitatea mediilor