Ordonati crescator !

[Marius Mainea]

Fie \( a_1,a_2,...,a_n,b_1,b_2,...,b_n \) numere pozitive astfel incat \( a_k^2-\frac{1}{k}=4b_k \) , \( k=\overline{1,n} \) , n natural fixat.
Ordonati crescator numerele \( A=\sum_{k=1}^{n} {\frac{a_k^2+b_k}{4a_k\sqrt{b_k}}} \) , \( B=\sum_{k=1}^{n} {\frac{\sqrt{b_k}}{a_k}} \) , \( C=\sum_{k=1}^{n} {\frac{a_k}{4\sqrt{b_k}}} \) si \( \frac{n}{2} \) .

,,Gh. Titeica'' 1999

[Marius Mainea]

Indicatie: \( B\le \frac{n}{2}\le C\le A \)

[Andi Brojbeanu]

\( B=\sum_{k=1}^n {\frac{\sqrt{b_k}}{a_k}}=\sum_{k=1}^n {\frac{\sqrt{\frac{a_k^2-\frac{1}{k}}{4}}}{a_k}}=\sum_{k=1}^n {\frac{\sqrt{a_k^2-\frac{1}{k}}}{2a_k}}\le \sum_{k=1}^n {\frac{\sqrt{a_k^2}}{2a_k}}=\sum_{k=1}^n {\frac{a_k}{2a_k}}=\frac{1}{2}\cdot n=\frac{n}{2}\Rightarrow B\le \frac{n}{2} \).

\( C=\sum_{k=1}^n {\frac{a_k}{4\sqrt{b_k}}}=\sum_{k=1}^n {\frac{a_k}{4 \sqrt{\frac{a_k^2-\frac{1}{k}}{4}}}}=\sum_{k=1}^n {\frac{a_k}{2\sqrt{a_k^2-\frac{1}{k}}}}\ge \sum_{k=1}^n {\frac{a_k}{2\sqrt{a_k^2}}}=\sum_{k=1}^n {\frac{a_k}{2a_k}}=\frac{1}{2}\cdot n= \frac{n}{2}\Rightarrow C\ge \frac{n}{2} \).


\( \frac{a_k^2+b_k}{4a_k\sqrt{b_k}}\ge \frac{a_k}{4\sqrt{b_k}} \Leftrightarrow a_k^2+b_k\ge a_k^2 \Leftrightarrow b_k\ge 0 \), adevarat, \( \forall k\in \mathb{N} \). Asadar, avem: \( \sum_{k=1}^n {\frac{a_k^2+b_k}{4a_k\sqrt{b_k}}}\ge \sum_{k=1}^n {\frac{a_k}{4\sqrt{b_k}}} \Rightarrow A\ge C \).

Deci, ordinea este \( B\le \frac{n}{2}\le C \le A \).

[mihai++]

O echivalenta nu este adevarata cand treci de la suma la termenul sumei, dar rezolvarea e foarte buna.

[Andi Brojbeanu]

Multumesc pentru sesizare. Am modificat si sper ca acum sa fie bine.