Algebra: Prove inequality $\frac{ab + bc + ca}{4S}\ge \operatorname{ctg} \frac{\pi}{6}$

Question

Prove this inequality where $a$, $b$ and $c$ are sides of a triangle and $S$ its Area. $$\frac{ab + bc + ca}{4S}\ge \operatorname{ctg} \frac{\pi}{6}$$

Answer 1

By the sine theorem, the given inequality is equivalent to $$ \frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C} \geq 2\sqrt{3} \tag{1}$$ and since $\frac{1}{\sin(x)}$ is a convex function on the interval $(0,\pi)$, $(1)$ is a straightforward consequence of Jensen's inequality.

Answer 2

Since $ab+ac+bc\geq\sum\limits_{cyc}(2ab-a^2)$ it's just $\sum\limits_{cyc}(a-b)^2\geq0$ and $$\sum_{cyc}(2ab-a^2)=\sum_{cyc}a(b+c-a)>0,$$ it's enough to prove that $$\sum_{cyc}(2ab-a^2)\geq4\sqrt3S$$ or $$\sum_{cyc}(2ab-a^2)\geq\sqrt{3\sum_{cyc}(2a^2b^2-a^4)}$$ or $$\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)\geq0,$$ which is Schur.

Done!