500 followers problem-1

Let $a^2=\tan(\alpha),b^2=\tan(\beta),c^2=\tan(\gamma),d^2=\tan(\delta)$ .Then $\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma)+\cos^2(\delta)=1$ .

Now, by the AM-GM inequality, $\sin^2(\alpha)=\cos^2(\beta)+\cos^2(\gamma)+\cos^2(\delta) \ge 3(\cos(\beta)\cos(\gamma)\cos(\delta))^{2/3}$ .

Multiplying this and other similar inequalities we get, $\sin^2(\alpha)\sin^2(\beta)\sin^2(\gamma)\sin^2(\delta) \ge 81 \cos^2(\alpha)\cos^2(\beta)\cos^2(\gamma)\cos^2(\delta)$ or $\tan(\alpha)\tan(\beta)\tan(\gamma)\tan(\delta) \ge 9$ .

Finally, $abcd=\sqrt{\tan(\alpha)\tan(\beta)\tan(\gamma)\tan(\delta)} \ge \color{#D61F06}{\boxed{3}}$ .

It can also be done using AM-GM only.

We have, $\dfrac{1}{1+a^4}+\dfrac{1}{1+b^4}+\dfrac{1}{1+c^4}+\dfrac{1}{1+d^4}=1$

So, from here, we get the following four things :

$\dfrac{1}{1+a^4}+\dfrac{1}{1+b^4}+\dfrac{1}{1+c^4}=\dfrac{d^4}{1+d^4}$ ..... $(1)$

$\dfrac{1}{1+a^4}+\dfrac{1}{1+b^4}+\dfrac{1}{1+d^4}=\dfrac{c^4}{1+c^4}$ ..... $(2)$

$\dfrac{1}{1+a^4}+\dfrac{1}{1+c^4}+\dfrac{1}{1+d^4}=\dfrac{b^4}{1+b^4}$ ..... $(3)$

$\dfrac{1}{1+b^4}+\dfrac{1}{1+c^4}+\dfrac{1}{1+d^4}=\dfrac{a^4}{1+a^4}$ ..... $(2)$

So, applying AM-GM on each of the statements $(1), (2), (3), (4)$ , we have :

$\dfrac{d^4}{1+d^4} \geq \dfrac{3}{\sqrt[3]{1+a^4} \sqrt[3]{1+b^4} \sqrt[3]{1+c^4}}$ ... $(A)$

$\dfrac{c^4}{1+c^4} \geq \dfrac{3}{\sqrt[3]{1+a^4} \sqrt[3]{1+b^4} \sqrt[3]{1+d^4}}$ ... $(B)$

$\dfrac{b^4}{1+b^4} \geq \dfrac{3}{\sqrt[3]{1+a^4} \sqrt[3]{1+c^4} \sqrt[3]{1+d^4}}$ ... $(C)$

$\dfrac{a^4}{1+a^4} \geq \dfrac{3}{\sqrt[3]{1+b^4} \sqrt[3]{1+c^4} \sqrt[3]{1+d^4}}$ ... $(D)$

Multiplying $(A), (B), (C), (D)$ , we get :

$(abcd)^4 \geq 3^4$

$\Rightarrow abcd \geq 3$

Can also be done by putting a=b=c=d.