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[不等式] 怎么用不等式法求证等式

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Canhuang Posted 2023-3-11 21:00 From mobile phone |Read mode
$\frac{\sin^4A}{\sin^2B}+\frac{\cos^4A}{\cos^2B}=1$,证明$\sin^2A=\sin^2B$.

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kuing Posted 2023-3-11 21:33
很简单啊,由均值
\begin{align*}
\frac{\sin^4A}{\sin^2B}+\sin^2B&\geqslant2\sin^2A,\\
\frac{\cos^4A}{\cos^2B}+\cos^2B&\geqslant2\cos^2A,
\end{align*}
相加结合条件变成 `1+1\geqslant2`,那只能取等,即得 `\sin^2A=\sin^2B`。

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isee Posted 2023-3-11 21:46
由Cauchy 不等式分式形式\[1=\frac{\big(\sin^2A\big)^2}{\sin^2B}+\frac{\big(\cos^2A\big)^2}{\cos^2B}\geqslant \frac{\big(\sin^2A+\cos^2A\big)^2}{\sin^2B+\cos^2B}=1,\]
于是仅等号成立\[\frac{\sin^2A}{\sin^2B}=\frac{\cos^2A}{\cos^2B}=\frac{\sin^2A+\cos^2B}{\sin^2A+\cos^2B}=1.\]
isee=freeMaths@知乎

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kuing Posted 2023-3-11 22:31
isee 发表于 2023-3-11 21:46
由Cauchy 不等式分式形式\[1=\frac{\big(\sin^2A\big)^2}{\sin^2B}+\frac{\big(\cos^2A\big)^2}{\cos^2B}\g ...
我第一反应也是这个,后来觉得还是用均值“初级”一点更容易被接受😏

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