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Fie \( a_1,a_2,...,a_n \) numere reale astfel incat \( 1 > {a_1} > {a_2} > {a_3} > ... > {a_n} > 0 \). Sa se demonstreze ca :

\( \frac {a^2_1}{1 - a_1} + \frac {a^2_2}{a_1 - a_2} + \frac {a^2_3}{a_2 - a_3} + ... + \frac {a^2_n}{a_{n - 1} - a_n} > \frac {1}{2}(a_1 + 2a_2 + 3a_3 + ... + na_n) - 1 \)

Inegalitatea este echivalenta cu


\( LHS+\sum_{k=1}^n \frac{(k+1)^2(a_{k-1}-a_k)}{4}> RHS +\sum_{k=1}^n \frac{(k+1)^2(a_{k-1}-a_k)}{4} \) unde \( a_0=1 \)

Insa aplicand AM-GM \( LHS+\sum_{k=1}^n \frac{(k+1)^2(a_{k-1}-a_k)}{4}=\sum_{k=1}^n \frac{a_k^2}{a_{k-1}-a_k}+\frac{(k+1)^2(a_{k-1}-a_k)}{4}\ge \sum_{k=1}^n(k+1)a_k>RHS +\sum_{k=1}^n \frac{(k+1)^2(a_{k-1}-a_k)}{4} \) ultima inegalitate fiind echivalenta cu

\( \frac{1}{4}a_1+\frac{1}{4}a_2+...+\frac{1}{4}a_{n-1}+\frac{n^2+4n+5}{4}a_n>0 \)