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Prove the inequality $$\tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}}+\tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}} > 1$$ for any $x, \alpha$ with $0 \leq x \leq \dfrac{\pi}{2}$ and $\dfrac{\pi}{6} < \alpha < \dfrac{\pi}{3}$.

The best idea I had was to use the identity $\tan(A+B) = \dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$. Thus we may say that $\tan \left({\dfrac{\pi \sin{x}}{4 \sin{\alpha}}}+\dfrac{\pi\cos{x}}{4\cos{\alpha}} \right) \left(1-\tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}} \tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}} \right) = \tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}}++\tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}}.$ Then I just have to show its greater than $1$ on these intervals.

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Since $\dfrac{\pi}{6} < \alpha < \dfrac{\pi}{3}$, we have $\sin \alpha > \dfrac{1}{2}$ and $\cos \alpha > \dfrac{1}{2}$.

Since $0 \le x \le \dfrac{\pi}{2}$, we have $0 \le \sin x \le 1$ and $0 \le \cos x \le 1$.

Hence, $0 \le \dfrac{\pi\sin x}{4\sin \alpha} < \dfrac{\pi}{2}$ and $0 \le \dfrac{\pi\cos x}{4\cos \alpha} < \dfrac{\pi}{2}$.

Therefore, $\tan \dfrac{\pi\sin x}{4\sin \alpha} \ge 0$ and $\tan \dfrac{\pi\cos x}{4\cos \alpha} \ge 0$ for all $0 \le x \le \dfrac{\pi}{2}$ and $\dfrac{\pi}{6} < \alpha < \dfrac{\pi}{3}$.

Now, we beak the proof into three cases:

Case I: If $x > \alpha$, then $\sin x > \sin \alpha$. Hence, $\tan \dfrac{\pi\sin x}{4\sin \alpha} > \tan\dfrac{\pi}{4} = 1$.

Therefore, $\tan \dfrac{\pi\sin x}{4\sin \alpha} + \tan \dfrac{\pi\cos x}{4\cos \alpha} > 1 + 0 = 1$.

Case II: If $x < \alpha$, then $\cos x > \cos \alpha$. Hence, $\tan \dfrac{\pi\cos x}{4\cos \alpha} > \tan\dfrac{\pi}{4} = 1$.

Therefore, $\tan \dfrac{\pi\sin x}{4\sin \alpha} + \tan \dfrac{\pi\cos x}{4\cos \alpha} > 0 + 1 = 1$.

Case III: If $x = \alpha$, then $\sin x = \sin \alpha$ and $\cos x = \cos \alpha$.

Therefore, $\tan \dfrac{\pi\sin x}{4\sin \alpha} + \tan \dfrac{\pi\cos x}{4\cos \alpha} = \tan \dfrac{\pi}{4} + \tan \dfrac{\pi}{4} = 1+1 = 2 > 1$.

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  • $\begingroup$ @user236182 What about the $x \geq \alpha$ case? $\endgroup$ – user19405892 Jan 11 '16 at 4:12
  • $\begingroup$ In the $x \geq \alpha$ case notice that it isn't strict. $\endgroup$ – user19405892 Jan 11 '16 at 4:14
  • $\begingroup$ Let me try to patch this up. $\endgroup$ – JimmyK4542 Jan 11 '16 at 4:14
  • $\begingroup$ @user236182 I don't believe that is necessarily true that if $A \geq 1$ then $B > 1$. $\endgroup$ – user19405892 Jan 11 '16 at 4:17
  • $\begingroup$ Okay, great that makes sense now! $\endgroup$ – user19405892 Jan 11 '16 at 4:18
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We could also do this without casework using Jensen inequality, and using the fact that the first derivative of $\tan(x)$ is always positive and the second derivative is positive when $0 < tan(x) < \frac{\pi}{2}$.

So, by Jensen we have,

$\tan{\left(\dfrac{\pi\sin{x}}{4\sin{\alpha}}\right)}+\tan{\left( \dfrac{\pi\cos{x}}{4\cos{\alpha}}\right)} \geq 2 \tan{\frac{\pi}{8} \left(\frac{\sin{x}}{\sin{\alpha}}+\frac{\cos{x}}{\cos{\alpha}}\right) } =2\tan{\frac{\pi}{8} \left(\frac{\sin(x+ \alpha)}{\sin(\alpha)\cos{\alpha}}\right)} \geq 2\tan{\frac{\pi}{8}\left(\frac{1}{cos \alpha}\right)} \geq 2\tan {\frac{\pi}{8}\left(\frac{1}{\frac{1}{2}}\right)} \geq 2$.

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