Câu hỏi 52
Nếu $a_i \ge 0$, hãy chứng minh rằng $$
\frac{n(n+1)}{2} \left( \prod_{i=1}^{n} A_i^i \right)^{\frac{2}{n(n+1)}} \ge \frac{n(n-1)}{2} \left( \prod_{i=1}^{n-1} A_i^i \right)^{\frac{2}{n(n-1)}} + nG_n,
$$ nơi $A_k = \sum_{i=1}^{k} \frac{2ia_i}{k(k+1)}$, $G_k = \left( \prod_{i=1}^{k} a_i^i \right)^{\frac{2}{k(k+1)}}$.
\frac{n(n+1)}{2} \left( \prod_{i=1}^{n} A_i^i \right)^{\frac{2}{n(n+1)}} \ge \frac{n(n-1)}{2} \left( \prod_{i=1}^{n-1} A_i^i \right)^{\frac{2}{n(n-1)}} + nG_n,
$$ nơi $A_k = \sum_{i=1}^{k} \frac{2ia_i}{k(k+1)}$, $G_k = \left( \prod_{i=1}^{k} a_i^i \right)^{\frac{2}{k(k+1)}}$.