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三角不等式/极值问题 - 语时lab
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<h1>三角不等式/极值问题</h1>
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<span class="date">2018/06/13</span>
<span>posted in </span>
<span class="posted-in"><a href='problemsolving.html'>解题</a></span>
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<p><mark>TODO: 本题确认伪证了</mark></p>
<p>已知 \(A,B,C\) 为三角形三内角. 求 \(\dfrac{\cos^2 A}{1+\cos A} + \dfrac{\cos^2 B}{1+\cos B} + \dfrac{\cos^2 C}{1+\cos C}\) 的最小值.</p>
<span id="more"></span><!-- more -->
<blockquote>
<p>Qer: zhangboxin 20180612</p>
</blockquote>
<h2 id="toc_0">解答</h2>
<p>设 \(t_1 = \tan \dfrac A2\), \(t_2 = \tan \dfrac B2\), \(t_3 = \tan \dfrac C2\). 则由 \(A + B + C = \pi\) 可导出<br/>
\[<br/>
t_1 t_2 + t_2 t_3 + t_3 t_1 = 1<br/>
\]<br/>
有<br/>
\[<br/>
1 + t_1^2 = t_1 t_2 + t_2 t_3 + t_3 t_1 + t_1^2 = (t_1+t_2)(t_1+t_3)<br/>
\]<br/>
以及<br/>
\[<br/>
t_1^2 + t_2^2 + t_3^2 = (t_1 + t_2 + t_3)^2 - 2<br/>
\]<br/>
又由 \(t_1^2 + t_2^2 + t_3^2 \ge \dfrac13 (t_1 + t_2 + t_3)^2\) 知<br/>
\[<br/>
(t_1 + t_2 + t_3)^2 \ge 3<br/>
\]<br/>
当且仅当 \(t_1 = t_2 = t_3 = \dfrac{\sqrt3}3\) 时取等号.</p>
<p>由万能公式<br/>
\[<br/>
\dfrac{\cos^2 A}{1 + \cos A} = \dfrac{(1-t_1^2)^2}{2(1+t_1^2)} = \dfrac{t_1^2 - 3}2 + \dfrac2{t_1^2 + 1} = \dfrac{t_1^2}2 + \dfrac2{(t_1+t_2)(t_1+t_3)} - \dfrac32<br/>
\]<br/>
从而<br/>
\[<br/>
\begin{aligned}<br/>
\text{原式} <br/>
&= \dfrac{t_1^2 + t_2^2 + t_3^2}2 + \dfrac{2[(t_2+t_3)+(t_3+t_1)+(t_1+t_2)]}{(t_1+t_2)(t_2+t_3)(t_3+t_1)} - \dfrac92 \\<br/>
&= \dfrac{(t_1 + t_2 + t_3)^2}2 + \dfrac{4(t_1+t_2+t_3)}{(t_1+t_2)(t_2+t_3)(t_3+t_1)} - \dfrac{11}2 \\<br/>
\end{aligned}<br/>
\]<br/>
由于<br/>
\[<br/>
\begin{aligned}<br/>
(t_1+t_2)(t_2+t_3)(t_3+t_1) <br/>
&\le \left[ \dfrac{(t_1+t_2)+(t_2+t_3)+(t_3+t_1)}3 \right]^3 \\<br/>
& = \dfrac8{27} (t_1+t_2+t_3)^3<br/>
\end{aligned}<br/>
\]<br/>
当且仅当 \(t_1 = t_2 = t_3\) 时取等号. 所以<br/>
\[<br/>
\text{原式} \ge \dfrac{(t_1 + t_2 + t_3)^2}2 + \dfrac{27}{2(t_1+t_2+t_3)^2} - \dfrac{11}2<br/>
\]<br/>
又由于 \((t_1 + t_2 + t_3)^2 \ge 3\), 故<br/>
\[<br/>
\text{原式} \ge \dfrac32 + \dfrac{27}{2\times3} - \dfrac{11}2 = \dfrac12<br/>
\]<br/>
当且仅当 \(t_1 = t_2 = t_3 = \dfrac{\sqrt3}3\) 时取等号, 此时 \(A = B = C = \dfrac\pi3\).</p>
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