For an acute angled triangle $ABC,$ if $p=\frac{\sqrt3+\sin A+\sin B+\sin C}{2\sin A\sin B\sin C}$, find the range of $p$ $$ p=\frac{\sqrt3+\sin A+\sin B+\sin C}{2\sin A\sin B\sin C}$$
$\displaystyle \sin A+\sin B+\sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$ and
$\displaystyle \sin A\sin B\sin C=8\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$ in a triangle
$A<90^\circ, \text{so} \space A/2<45^\circ$ then $ \sin\frac{A}{2}<1/\sqrt2$. However, $$ \cos\frac{A}{2}>1/\sqrt2$$
So this probably doesn't lead us anywhere :( .
Can anyone please help, thanks.
 A: The lower bound.
For $\alpha=\beta=\gamma=60^{\circ}$ we obtain a value $\frac{10}{3}$.
We'll prove that it's a minimal value.
Indeed, in the standard notation we need to prove that $$\sqrt3+\sum_{cyc}\frac{2S}{bc}\geq\frac{20}{3}\cdot\frac{8S^3}{a^2b^2c^2}$$ or
$$\sqrt3+\frac{2S(a+b+c)}{abc}\geq\frac{160S^3}{3a^2b^2c^2}.$$
Here $S$ is the area of the triangle
Now, let $a=\frac{y+z}{2}$, $b=\frac{x+z}{2}$ and  $c=\frac{x+y}{2}$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that:
$$\sqrt3+\frac{4\sqrt{xyz(x+y+z)^3}}{\prod\limits_{cyc}(x+y)}\geq\frac{160\sqrt{x^3y^3z^3(x+y+z)^3}}{3\prod\limits_{cyc}(x+y)^2}$$ or
$$\prod_{cyc}(x+y)+12\sqrt{xyz\left(\frac{x+y+z}{3}\right)^3}\geq\frac{160\sqrt{x^3y^3z^3(x+y+z)^3}}{3\sqrt3\prod\limits_{cyc}(x+y)}.$$
Now, by AM-GM we obtain:
$$\prod_{cyc}(x+y)+12\sqrt{xyz\left(\frac{x+y+z}{3}\right)^3}=2\cdot\frac{\prod\limits_{cyc}(x+y)}{2}+3\cdot4\sqrt{xyz\left(\frac{x+y+z}{3}\right)^3}\geq$$
$$\geq5\sqrt[5]{\left(\frac{\prod\limits_{cyc}(x+y)}{2}\right)^2\left(4\sqrt{xyz\left(\frac{x+y+z}{3}\right)^3}\right)^3}$$ and it's enough to prove that:
$$5\sqrt[5]{\left(\frac{\prod\limits_{cyc}(x+y)}{2}\right)^2\left(4\sqrt{xyz\left(\frac{x+y+z}{3}\right)^3}\right)^3}\geq\frac{160\sqrt{x^3y^3z^3(x+y+z)^3}}{3\sqrt3\prod\limits_{cyc}(x+y)}$$ or
$$3^6\prod_{cyc}(x+y)^{14}\geq2^{42}(xyz)^{12}(x+y+z)^6.$$
Now, use $$\prod_{cyc}(x+y)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)\geq\frac{8}{9}(x+y+z)\sqrt{3xyz(x+y+z)}$$ and $$x+y+z\geq3\sqrt[3]{xyz}.$$
The upper bound is $+\infty$.
Try $\alpha=\beta\rightarrow\left(90^{\circ}\right)^-$ and $\gamma\rightarrow\left(0^{\circ}\right)^+$.
