设 $0<\alpha,\beta,\gamma<\dfrac{\pi}{2}$，且 $\sin^3\alpha+\sin^3\beta+\sin^3\gamma=1$，求证：$\tan^2\alpha+\tan^2\beta+\tan^2\gamma\geqslant\dfrac{3\sqrt{3}}{2}$．
【难度】
【出处】
2005中国东南数学奥林匹克试题
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【答案】
略
【解析】
令 $a=\sin\alpha,b=\sin\beta,c=\sin\gamma$，则 $a,b,c\in(0,1)$，且 $a^3+b^3+c^3=1$，
$\begin{aligned}
a-a^3&=\dfrac{1}{\sqrt{2}}\sqrt{2a^2(1-a^2)^2}\\
&\leqslant \dfrac{1}{\sqrt{2}}\sqrt{\left(\dfrac{2a^2+1-a^2+1-a^2}{3}\right)^3}\\
&=\dfrac{2}{3\sqrt{3}}\end{aligned}$
同理 $b-b^3\leqslant\dfrac{2}{3\sqrt{3}},c-c^2\dfrac{2}{3\sqrt{3}}$
因此
$\begin{aligned}
&\dfrac{a^2}{1-a^2}+\dfrac{b^2}{1-b^2}+\dfrac{c^2}{1-c^2}\\
=&\dfrac{a^3}{a-a^3}+\dfrac{b^3}{b-b^3}+\dfrac{c^3}{c-c^3}\\
\geqslant&\dfrac{3\sqrt{3}}{2}(a^3+b^3+c^3)=\dfrac{3\sqrt{3}}{2}
\end{aligned}$
注意到
$\tan^2\alpha=\dfrac{\sin^2\alpha}{1-\sin^2\alpha}=\dfrac{a^2}{1-a^2}$
$\tan^2\beta=\dfrac{\sin^2\beta}{1-\sin^2\beta}=\dfrac{b^2}{1-b^2}$
$\tan^2\gamma=\dfrac{\sin^2\gamma}{1-\sin^2\gamma}=\dfrac{c^2}{1-c^2}$
所以 $\tan^2\alpha+\tan^2\beta+\tan^2\gamma\geqslant\dfrac{3\sqrt{3}}{2}$．
$\begin{aligned}
a-a^3&=\dfrac{1}{\sqrt{2}}\sqrt{2a^2(1-a^2)^2}\\
&\leqslant \dfrac{1}{\sqrt{2}}\sqrt{\left(\dfrac{2a^2+1-a^2+1-a^2}{3}\right)^3}\\
&=\dfrac{2}{3\sqrt{3}}\end{aligned}$
同理 $b-b^3\leqslant\dfrac{2}{3\sqrt{3}},c-c^2\dfrac{2}{3\sqrt{3}}$
因此
$\begin{aligned}
&\dfrac{a^2}{1-a^2}+\dfrac{b^2}{1-b^2}+\dfrac{c^2}{1-c^2}\\
=&\dfrac{a^3}{a-a^3}+\dfrac{b^3}{b-b^3}+\dfrac{c^3}{c-c^3}\\
\geqslant&\dfrac{3\sqrt{3}}{2}(a^3+b^3+c^3)=\dfrac{3\sqrt{3}}{2}
\end{aligned}$
注意到
$\tan^2\alpha=\dfrac{\sin^2\alpha}{1-\sin^2\alpha}=\dfrac{a^2}{1-a^2}$
$\tan^2\beta=\dfrac{\sin^2\beta}{1-\sin^2\beta}=\dfrac{b^2}{1-b^2}$
$\tan^2\gamma=\dfrac{\sin^2\gamma}{1-\sin^2\gamma}=\dfrac{c^2}{1-c^2}$
所以 $\tan^2\alpha+\tan^2\beta+\tan^2\gamma\geqslant\dfrac{3\sqrt{3}}{2}$．
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