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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.

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Those arguments are in degrees, no? –  J. M. Aug 9 '12 at 16:12
    
Yup degrees only. –  TheApe Aug 9 '12 at 16:12
    
It seems like you can do standard nonlinear optimization, like Lagrange multipliers, have you tried this? –  Cocopuffs Aug 9 '12 at 16:13
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$\frac{\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ}}{4} ≥ (p^2q^2r^2s^2)^\frac{1}{4}$ as $\tan9^\circ.\tan81^\circ=1$ and $\tan27^\circ.\tan63^\circ=1$ $\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ} ≥4\sqrt{pqrs}$ –  lab bhattacharjee Aug 9 '12 at 16:28
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${\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ}}$ $={\frac{p^2}{\tan9^\circ} + \frac{s^2}{\tan81^\circ}} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ}$ $=2(ps+rs)$ as $\tan9^\circ.\tan81^\circ=1$ and $\tan27^\circ.\tan63^\circ=1$ –  lab bhattacharjee Aug 9 '12 at 16:28
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1 Answer

up vote 5 down vote accepted

$(\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ) \left ( \frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ} \right ) \geq {(p+q+r+s)}^2=5^2$

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

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Thank you. Got it. –  TheApe Aug 9 '12 at 16:49
    
@TheApe glad I could help. –  clark Aug 9 '12 at 17:51
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