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.
What will be the minimum value of $\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ}$?
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1${\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ}}$ $={\frac{p^2}{\tan9^\circ} + \frac{s^2}{\tan81^\circ}} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ}$ $=2(ps+rs)$ as $\tan9^\circ.\tan81^\circ=1$ and $\tan27^\circ.\tan63^\circ=1$ – 2012-08-09
1 Answers
$(\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ) \left ( \frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ} \right ) \geq {(p+q+r+s)}^2=5^2$
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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1@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