If $x + y + z + w=5$ then the minimum value of $x^2 \cot (9°) + y^2 \cot (27°) + z^2 \cot (63°) + w^2 \cot (81°)$ is?

Question

I used the $A.M. \ge G.M.$ rule.

$$((xyzw)^2 . \cot(9^o)\tan(9^o)\cot(27^o)\tan(27^o))^{1/4} \le \space \frac {x^2 \cot (9°) + y^2 \cot (27°) + z^2 \cot (63°) + w^2 \cot (81°)}{4}$$

This reduces to : $$4(xyzw)^{1/2} \le x^2 \cot (9°) + y^2 \cot (27°) + z^2 \cot (63°) + w^2 \cot (81°) ---- (1)$$

Now, I applied the same thing to the first equation:

$$ \frac {x+y+z+w}{4} \ge (xywz)^{1/4}$$ $$ \frac {5}{4} \ge (xywz)^{1/4}$$ $$ \frac {25}{16} \ge (xywz)^{1/2} ----(2)$$

Now I do not know what to do. From equation (2), I get the minimum value of $(xywz)^{1/2}$ as $- \infty$ (Am I right over here ?). In a nutshell, I am confused what to do next.

Answer 1

By C-S $$\left(x^2 \cot 9° + y^2 \cot27° + z^2 \cot63° + w^2 \cot 81°\right)\left(\tan9^{\circ}+ \tan27^{\circ}+\tan63^{\circ}+\tan81^{\circ}\right)\geq(x+y+z+w)^2=25$$ The equality occurs for $$\left(x\sqrt{ \cot 9°}, y\sqrt{ \cot27°}, z\sqrt{ \cot63°},w\sqrt{ \cot 81°}\right)||\left(\sqrt{\tan9^{\circ}}, \sqrt{\tan27^{\circ}},\sqrt{\tan63^{\circ}},\sqrt{\tan81^{\circ}}\right),$$ which gives the answer: $$\frac{25}{\tan9^{\circ}+ \tan27^{\circ}+\tan63^{\circ}+\tan81^{\circ}}.$$ Done!