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Nu o sa postez decat o problema intrucat celelalte sunt cam usoare.

Problema. Fie \( (a_{n})_{n\geq 1} \) un sir de numere reale strict pozitive, pentru care: \( a_{n}^2<a_{n}-a_{n+1},\ (\forall)n\in\mathbb{N}^*. \)

Sa se arate ca: \( \lfloor n\cdot a_{n}\rfloor=0,\ (\forall)n\in\mathbb{N}^*. \)

Iuliana Duma

O sa o fac cu inductie :
\( p(n): a_n<\frac{1}{n} \).
\( p(1) \) e adevarata caci \( 0<a_2<a_1(1-a_1)\Rightarrow1>a_1. \)
\( a_2<a_1(1-a_1)<\frac{1}{2}. \)
\( a_n^2<a_n-a_{n+1}\Rightarrow a_{n+1}<a_n-a_n^2,n\geq2 \) si cum functia \( f(x)=x-x^2 \) isi atinge maximul in \( \frac{1}{n} \) caci e strict crescatoare pana la \( \frac{1}{2} \) si descrescatoare de la \( \frac{1}{2} \), avem ca
\( a_{n+1}<\frac{1}{n}-\frac{1}{n^2}<\frac{1}{n+1} \). Ultima inegalitate fiind echivalenta cu \( n^2>(n-1)(n+1) \).
In concluzie \( na_n<1 \Rightarrow [na_n]=0,\forall n \).

Metoda 2

\( a_n^2 < a_n - a_{n+1} => a_n(a_n-1) < -a_{n+1} < 0 => a_n \in (0,1) \)

Cum \( a_n \in (0,1) => \frac{1}{a_n(1-a_n)} < \frac{1}{a_{n+1}} => \frac{1}{a_n} + \frac{1}{1-a_n} < \frac{1}{a_{n+1}} => \frac{1}{1-a_n} < \frac{1}{a_{n+1}} - \frac{1}{a_n} \)
\( => \sum_{k=1}^{n-1}\frac{1}{1-a_n} < \sum_{k=1}^{n-1}\(\frac{1}{a_{n+1}} - \frac{1}{a_n}\) => \frac{1}{1-a_1} + \frac{1}{1-a_2} + ... + \frac{1}{1-a_{n-1}} < \frac{1}{a_n} - \frac{1}{a_1} . \)

Folosim \( 1-a_i \in (0,1) => \frac{1}{1-a_i}>1 \) oricare ar fi \( i \to n-1 \)

\( a_1 \in (0,1) => \frac{1}{a_1} > 1. \)

\( \frac{1}{a_n} > (n-1)+1 = n => na_n < 1 \)

Cum \( na_n > 0 =>[na_n] = 0 \).