<s>[INST] Showing $\sum_{cyc}\tan\frac\alpha2\tan\frac\beta2\geq4$ for a cyclic quadrilateral 
Let $ABCD$ be a cyclic quadrilateral with sides $a$, $b$, $c$ and $d$. Denote $s$ the semiperimeter and let $\angle{DAB}=\alpha$, $\angle{ABC}=\beta$, $\angle{BCD}=\gamma$ and $\angle{CDA}=\delta$. Then the following inequality holds
$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}\geq4.\tag{1}$$

Proof. Substituting from the half-angle formula for the tangent (see $(2)$ here) we have that
$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}=\sqrt{\frac{(s-a)(s-d)}{(s-b)(s-c)}}\cdot{\sqrt{\frac{(s-a)(s-b)}{(s-c)(s-d)}}}=\frac{s-a}{s-c}.$$
Similarly,
$$\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=\frac{s-b}{s-d}\qquad\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}=\frac{s-c}{s-a}\qquad\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}=\frac{s-d}{s-b}$$
Thus, the left-hand side of $(1)$ can be rewritten as follows
$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}=\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}.$$
But,
$$\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}=\frac{a-c}{s-a}+\frac{b-d}{s-b}+\frac{c-a}{s-c}+\frac{d-b}{s-d}+4.$$
Since $\frac{a-c}{s-a}+\frac{c-a}{s-c}=\frac{4(a-c)^2}{(-a+b+c+d)(a+b-c+d)}$, and similarly for $\frac{b-d}{s-b}+\frac{d-b}{s-d}$, then $\frac{a-c}{s-a}+\frac{b-d}{s-b}+\frac{c-a}{s-c}+\frac{d-b}{s-d}$ is positive. Hence,

$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}\geq4.$$

Notice equality holds when $ABCD$ is rectangular.
A huge list of inequalities can be seen at Cut-the-knot.org.
Questions: a) Is there a simpler way to prove $(1)$? b) Is $(1)$ known?
 [/INST] Use AM_GM
$F=\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}\ge 4 \sqrt[4]{\frac{s-a}{s-c}\frac{s-b}{s-d}\frac{s-c}{s-a}\frac{s-d}{s-b}}=4.$
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