Cosec Raining (fixed)

Since $A\csc A$ , $B\csc B$ and $C\csc C$ are identical function, $y$ is maximum when $x\csc x$ is maximum. We shall prove that $f(x) = x\csc x = \dfrac x{\sin x}$ is an increasing function for $0 < x \le \frac \pi 2$ . Therefore, the maximum of $f(x)$ , $f_{max} = f \left(\frac \pi 2\right) = \dfrac {\frac \pi 2}{\sin \frac \pi 2} = \dfrac \pi 2$ . And the maximum pf $y$ , $y_{max} = 3f_{max} = \dfrac {3\pi}2 \approx \boxed{4.712}$ .

Proof: Let us prove that $f(x) = \dfrac x{\sin x}$ is an increasing function for $0 < x < \frac \pi 2$ . We need to prove that $f'(x) > 1$ .

$\begin{aligned} f(x) & = \frac x{\sin x} \\ f'(x) & = \frac d{dx}\left(\frac x{\sin x} \right) \\ & = \frac 1{\sin x} - \frac {x\cos x}{\sin^2 x} \\ & = \frac 1{\sin x}\left(1-\frac x{\color{#3D99F6}\tan x}\right) & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \frac 1{\sin x}\left(1- {\color{#3D99F6} \frac x{x+\frac {x^3}3 + \frac {2x^5}{15}+...}}\right) & \small \color{#3D99F6} \text{Dividing up and down by }x \\ & = \frac 1{\sin x}\left(1-{\color{#3D99F6} \frac 1{1+\frac {x^2}3 + \frac {2x^4}{15}+...}}\right) & \small \color{#3D99F6} \text{Note that } \frac 1{1+\frac {x^2}3 + \frac {2x^4}{15}+...} < 1 \end{aligned}$

We note that $\dfrac 1{\sin x} > 0$ and $1-\dfrac x{\tan x} > 0$ , $\implies f'(x) > 0$ and $f(x)$ , an increasing function, for $0 < x < \frac \pi 2$ .

$xcosec(x)$ is an increasing function. in (0,π/2).

Because ,

Consider $f(x)$ = $xcosec(x)$ ,

$f'(x)$ = $cosec(x) - xcosec(x)cot(x)$ ,

$f'(x)$ = $cosec(x)(1-xcot(x))$ ,

$f'(x)$ = $(cosec(x)(tan(x) - x)/tan(x)$ ,

We know that, $sin(x)$ < $x$ < $tan(x)$ .

So , we can conclude that $f'(x)$ is positive in(0,π/2).

So $f(x)$ is increasing function,

So it is maximum at π/2 ,

Therefore​, the given equation has maximum at A=B=C=π/2.

So maximum value is 3π/2 = 4.172 .