TITLE: What will be the minimum value of $\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ}$?
QUESTION [6 upvotes]: What will be the minimum value of
$$\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ}$$ if 
$$p+q+r+s=5$$ where $p, q, r, s$ are positive reals?
I tried applying AM-GM inequality but it didn't help.

REPLY [7 votes]: $(\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ) \left ( \frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ} \right ) \geq  {(p+q+r+s)}^2=5^2$ 
From Cauchy–Schwarz 
Hence
$$\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ} \, \, \, \,\geq \frac{  5^2 }{\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ }$$
Then take $$p = \frac{ 5 \tan9^\circ}{\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ}$$
$$q = \frac{ 5 \tan27^\circ}{\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ}$$
$$r = \frac{ 5 \tan63^\circ}{\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ}$$
$$s = \frac{ 5 \tan81^\circ}{\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ}$$
And sum up.
Note that those $ \tan  $ are positive