# 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 :( .

• Source of the question, please?
– Babu
Aug 31, 2021 at 5:12
• @Buraian, I found this in a jee advanced mock test en.wikipedia.org/wiki/Joint_Entrance_Examination_–_Advanced Aug 31, 2021 at 5:14
• Some rearranging gives $\displaystyle p=\frac{\sqrt 3}{8\Pi\cos\frac{ A}{2}(2\Pi\sin \frac{A}{2}-1)}$ Aug 31, 2021 at 5:21
• A way to solve this without much thought would be to use lagrange multipliers
– Babu
Aug 31, 2021 at 5:46
• Can you post a solution using Lagrange multipliers ? @Buraian Sep 9, 2021 at 8:56

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 $$\gamma\rightarrow\left(0^{\circ}\right)^+$$.

• What does S denote? Aug 31, 2021 at 7:43
• @Lalit Tolani It's an area of the triangle. Aug 31, 2021 at 7:45
• $\sqrt3+\frac{2S(a+b+c)}{abc}\geq\frac{160S^3}{3a^2b^2c^2}$, how did you simplify the original expression to this one, and are a,b,c the sides of the triangle Aug 31, 2021 at 8:29
• @Sunaina We know that $S=\frac{1}{2}bc\sin\alpha$, which gives $\sin\alpha=\frac{2S}{bc}.$ Aug 31, 2021 at 8:33
• @MichaelRozenberg, I've noticed something else (though not exactly related to this question), in many inequality questions (not all) we can get the minimum/maximum value if we substitute a=b=c=d.... (depending on the number of variables). Is there a theorem for this (i.e a function must satisfy certain properties for this to be true), or is it completely arbitrary and random. Aug 31, 2021 at 10:38