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          <h1>两个数列不等式问题</h1>
     
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          <span class="date">2019/02/01</span>

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              <span class="posted-in"><a href='problemsolving.html'>解题</a></span>
           
         
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      <ol>
<li>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.</li>
<li>设 \(a_n = 1 + \dfrac12 + \dfrac13 + \dots + \dfrac1n\). 求证: 对 \(n\ge 2\) 有 \(a_n^2 &lt; 2 \left( \dfrac{a_2}2 + \dfrac{a_3}3 + \dots \dfrac{a_n}n \right) + \dfrac{n+3}{2n+2}\).</li>
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<blockquote>
<p>Qer: xieshunfu 20190201</p>
</blockquote>

<h2 id="toc_0">解答</h2>

<h3 id="toc_1">题 1</h3>

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

<h3 id="toc_2">题 2</h3>

<blockquote>
<p>数学归纳法, 不难!</p>
</blockquote>


    

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