Câu hỏi 11
Cho $0 \lt a_1 \lt a_2 \lt \dots \lt a_n$ là các số thực. Chứng minh rằng
$$
\begin{aligned}
&(\frac{1}{1+a_1} + \frac{1}{1+a_2} + \dots + \frac{1}{1+a_n})^2 \le \frac{1}{a_1} + \frac{1}{a_2-a_1} + \frac{1}{a_3-a_2} \\
&+ \dots + \frac{1}{a_n-a_{n-1}}.
\end{aligned}
$$
$$
\begin{aligned}
&(\frac{1}{1+a_1} + \frac{1}{1+a_2} + \dots + \frac{1}{1+a_n})^2 \le \frac{1}{a_1} + \frac{1}{a_2-a_1} + \frac{1}{a_3-a_2} \\
&+ \dots + \frac{1}{a_n-a_{n-1}}.
\end{aligned}
$$