Minimum and Maximum range of $\frac{\cot A+\cot B}{\sin^2C} + \frac{2\cot B+2\cot C}{\sin^2 A} - \frac{3\cot C+3\cot A}{\sin^2 B} $

Question

Consider any acute triangle $ABC$ and given range

$\ \ x \leq \frac{\cot A+\cot B}{\sin^2 C} + \frac{2cotB+2cotC}{sin^2 A} - \frac{3cotC+3cotA}{sin^2 B} \leq y$

What is the value of $y-x$ ?

Basically , I have to confess that I cannot figured out solution to start with.

Only thing that I can do is to change $cot A + cot B , cot B + cot C ,cot C + cot A$ to another form.

We know that $A+B+C = \pi$ and the way I did is not solvable.

Thank ypu for any help.

Answer 1

Let $\cot\alpha=x$, $\cot\beta=y$ and $\cot\gamma=z$.

Thus, $xy+xz+yz=1$ and our expression it's: $$(x+y)(1+z^2)+2(y+z)(1+x^2)-3(x+z)(1+y^2)=$$ $$=(x+y)(x+z)(y+z)+2(y+z)(x+y)(x+z)-3(x+z)(x+y)(y+z)=0.$$

Answer 2

Hint

$$\dfrac{\cot A+\cot B}{\sin^2C}=\dfrac{\sin(A+B)}{\sin^2C\sin A\sin B}=\dfrac1{\sin A\sin B\sin C}$$

as $\sin(A+B)=\sin(\pi-C)=\sin C$

which will $>0$ for $0<A<\pi$

So, the condition of acute angle is not exactly necessary