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<h1>数列不等式 `lhc@jinan|20200407`</h1>
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<span class="date">2020/04/08</span>
<span>posted in </span>
<span class="posted-in"><a href='problemsolving.html'>解题</a></span>
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<p>已知数列 \(\{a_n\}\) 满足 \(a_1 = 1\), \(a_n a_{n+1} = n\), \(n = 1,2,3,\dots\). 求证: <br/>
\[<br/>
\frac1{a_1} + \frac1{a_2} + \frac1{a_3} + \dots + \frac1{a_n} \ge 2 \sqrt n - 1.<br/>
\]</p>
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<blockquote>
<p>Qer: lihaocheng@jinan 20200407</p>
</blockquote>
<h2 id="toc_0">解答</h2>
<p>这里给出一个解法, 过程相当简洁漂亮, 但有一个最大的问题就是, 似乎无法讲清楚这个思路是怎么来的, 这种方法是怎么想到的.</p>
<p>不难确认数列 \(\{a_n\}\) 的所有项都是正的, 由均值不等式<br/>
\[<br/>
\frac1{a_n} + \frac1{a_{n+1}} \ge 2 \sqrt{\frac1{a_n a_{n+1}}} = \frac2{\sqrt n}<br/>
\]<br/>
事实上, 这个等号也是取不到的, 不过这一点说起来比较费事, 不过后面可以看到也可以先不说这件事情.</p>
<p>于是, <strong>当 \(n \ge 2\) 时</strong>,<br/>
\[<br/>
\begin{aligned}<br/>
&<br/>
\frac1{a_1} + \frac1{a_2} + \frac1{a_3} + \dots + \frac1{a_n} \\<br/>
{}={}& <br/>
\frac12 \left[ <br/>
\frac1{a_1} + \left( \frac1{a_1} + \frac1{a_2} \right) + \left( \frac1{a_2} + \frac1{a_3} \right) + \dots + \left( \frac1{a_{n-1}} + \frac1{a_n} \right) + \frac1{a_n}<br/>
\right] \\<br/>
{}\ge{}&<br/>
\frac12 \left(<br/>
\frac1{a_1} + \frac2{\sqrt1} + \frac2{\sqrt2} + \dots + \frac2{\sqrt{n-1}} + \frac1{a_n}<br/>
\right) \\<br/>
{}={}&<br/>
\frac12 + \frac1{\sqrt1} + \frac1{\sqrt2} + \dots + \frac1{\sqrt{n-1}} + \frac1{2a_n}<br/>
\end{aligned}<br/>
\]</p>
<p>另一方面, 用数学归纳法不难证明<br/>
\[<br/>
\frac1{\sqrt1} + \frac1{\sqrt2} + \dots + \frac1{\sqrt{n-1}} \ge 2 \sqrt n - 1<br/>
\]</p>
<p>这样就不难证明原题中的不等式.</p>
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