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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?2012-08-09
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    Yup degrees only.2012-08-09
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    It seems like you can do standard nonlinear optimization, like Lagrange multipliers, have you tried this?2012-08-09
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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}$2012-08-09
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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$2012-08-09

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

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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    @TheApe glad I could help.2012-08-09
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    How did you find those values of p, q r and s? I understood upto the inequality statement.2017-12-17
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    @Rick The inequality is not enough to find the minimum of the expression, as the lower bound may not be attained for $p,q,r,s$ as described. So first we find the values such that the lower bounded is attained by knowing when C-S achieves its minimum. The C-S inequality $$(\sum_i a_i^2)(\sum_i b_i^2)\geq (\sum_i a_ib_i)^2$$ achieves its minimum when $a_i=\lambda b_i$ for some $\lambda. $ The latter justifies the choice for $p,q,r,s$. We chose $\lambda$ so that the sum of $p,q,r,s$ is $5$.2017-12-17