Prove the inequality $\tan{\frac{\pi\sin{x}}{4\sin{\alpha}}}+\tan{\frac{\pi\cos{x}}{4\cos{\alpha}}} > 1$ 
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.

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