1. 已知实数 \(x_1, x_2, \dots, x_{10}​\), \(\sum\limits_{k=1}^{10} k x_k = 1​\). 求 \(\left( \sum\limits_{k=1}^{10} x_k \right)^2 + \sum\limits_{k=1}^{10} x_k^2​\) 的最小值.
2. 设 \(a_n = 1 + \dfrac12 + \dfrac13 + \dots + \dfrac1n\). 求证: 对 \(n\ge 2\) 有 \(a_n^2 < 2 \left( \dfrac{a_2}2 + \dfrac{a_3}3 + \dots \dfrac{a_n}n \right) + \dfrac{n+3}{2n+2}\).

Qer: xieshunfu 20190201

解答

题 1

对于任意实数 \(t\), 均有 (Cauchy 不等式):
\[
\begin{aligned}
\left[ \left( \sum_{k=1}^{10} x_k \right)^2 + \sum_{k=1}^{10} x_k^2 \right]
\left[ t^2 + \sum_{k=1}^{10} (k-t)^2 \right]
&\ge
\left[ \left( \sum_{k=1}^{10} x_k \right) \cdot t + \sum_{k=1}^{10} (k-t) x_k \right]^2 \\
&= \left( \sum_{k=1}^{10} k x_k \right)^2 = 1
\end{aligned}
\]
即
\[
\left[ \left( \sum_{k=1}^{10} x_k \right)^2 + \sum_{k=1}^{10} x_k^2 \right]
\cdot (11 t^2 - 110 t + 385) \ge 1
\]
或
\[
\left( \sum_{k=1}^{10} x_k \right)^2 + \sum_{k=1}^{10} x_k^2
\ge \dfrac1{11 t^2 - 110 t + 385}
\]
此不等式对一切 \(t \in \mathbb R\) 都成立. 因此
\[
\left( \sum_{k=1}^{10} x_k \right)^2 + \sum_{k=1}^{10} x_k^2
\ge \left( \dfrac1{11 t^2 - 110 t + 385} \right)_{\max} =
\left( \dfrac1{11 t^2 - 110 t + 385} \right)_{t=5} = \dfrac1{110}
\]
当且仅当
\[
\begin{aligned}
\left( \sum_{k=1}^{10} x_k \right) : x_1 : x_2 : \dots : x_{10}
&= t : (1-t) : (2-t) : \dots : (10-t) \\
&= 5 : (-4) : (-3) : \dots : 5
\end{aligned}
\]
时取等号.

题 2

数学归纳法, 不难!