三角不等式

longzaifei

证明：$ \tan 0.5\du (\tan 0.5\du +\tan 1.5 \du +\dots+\tan 59.5 \du )>\dfrac{1}{3}  $

Tesla35

证：\[
\begin{aligned}
&\tan \left(0.5^{\circ}\right)\left(\tan \left(0.5^{\circ}\right)+\tan \left(1.5^{\circ}\right)+\tan \left(2.5^{\circ}\right)+\cdots \cdots+\tan \left(59.5^{\circ}\right)\right)\\
& =\frac{\sin 0.5^{\circ} \sin0.5^{\circ}}{\cos 0.5^{\circ} \cos 0.5^{\circ}}+\frac{\sin 0.5^{\circ} \sin 1.5^{\circ}}{\cos 0.5^{\circ} \cos 1.5^{\circ}}+\cdots \cdots+\frac{\sin 0.5^{\circ} \sin 59.5^{\circ}}{\cos 0.5^{\circ} \cos 59.5^{\circ}} \\
& =\frac{\cos 0^{\circ}-\cos 1^{\circ}}{\cos 0^{\circ}+\cos 1^{\circ}}+\frac{\cos 1^0-\cos 2^{\circ}}{\cos 1^{\circ}+\cos 2^{\circ}}+\cdots \cdots+\frac{\cos 59^{\circ}-\cos 60^{\circ}}{\cos 59^{\circ}+\cos 60^{\circ}} \\
& =\frac{\left(\cos 0^{\circ}-\cos 1^{\circ}\right)^2}{\cos ^2 0^{\circ}-\cos ^2 1^{\circ}}+\frac{\left(\cos 1^{\circ}-\cos 2^{\circ}\right)^2}{\cos ^2 1^{\circ}-\cos ^2 2^{\circ}}+\cdots \cdots+\frac{\left(\cos 59^{\circ}-\cos 60^{\circ}\right)^2}{\cos ^2 59^{\circ}-\cos ^2 60^{\circ}}
\end{aligned}
\]
根据柯西不等式得：
\[
\begin{aligned}
& \left(\frac{\left(\cos 0^{\circ}-\cos 1^{\circ}\right)^2}{\cos ^2 0^{\circ}-\cos ^2 1^{\circ}}+\frac{\left(\cos 1^{\circ}-\cos 2^{\circ}\right)^2}{\cos ^2 1^{\circ}-\cos ^2 2^{\circ}}+\cdots \cdots+\frac{\left(\cos 59^{\circ}-\cos 60^{\circ}\right)^2}{\cos ^2 59^{\circ}-\cos ^2 60^{\circ}}\right) \\
& \cdot\left(\left(\cos ^2 0^{\circ}-\cos ^2 1^{\circ}\right)+\left(\cos ^2 1^{\circ}-\cos ^2 2^{\circ}\right)+\cdots \cdots+\left(\cos ^2 59^{\circ}-\cos ^2 60^{\circ}\right)\right) \\
& >\left(\left(\cos 0^{\circ}-\cos 1^{\circ}\right)+\left(\cos 1^{\circ}-\cos 2^{\circ}\right)+\cdots \cdots\left(\cos 59^{\circ}-\cos 60^{\circ}\right)\right)^2 \\
& \text{即 }\left(\frac{\left(\cos 0^{\circ}-\cos 1^{\circ}\right)^2}{\cos ^2 0^{\circ}-\cos ^2 1^{\circ}}+\frac{\left(\cos 1^{\circ}-\cos 2^{\circ}\right)^2}{\cos ^2 1^{\circ}-\cos ^2 2^{\circ}}+\cdots \cdots+\frac{\left(\cos 59^{\circ}-\cos 60^{\circ}\right)^2}{\cos ^2 59^{\circ}-\cos ^2 60^{\circ}}\right) \\
& >\frac{\left(\cos 0^{\circ}-\cos 60^{\circ}\right)^2}{\cos ^2 0-\cos ^2 60}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}(\text{注 显然等号不能成立}) \\
& \therefore \tan \left(0.5^{\circ}\right)\left(\tan \left(0.5^{\circ}\right)+\tan \left(1.5^{\circ}\right)+\tan \left(2.5^{\circ}\right)+\cdots \cdots+\tan \left(59.5^{\circ}\right)\right)>\frac{1}{3}
\end{aligned}
\]

longzaifei

实在是太厉害了！！！！

其妙

回复 2# Tesla35
出处dang！

kuing

回复 2# Tesla35

题撸多了就是方便，585