In an acute angled triangle with angles $A,B$ and $C$, prove that $\left(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} \right)> \frac6\pi$ In an Acute Angled Triangle Prove That : -
$$\left(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} \right)> \frac{6}{\pi}$$
A , B , C represent the angles of the triangle.
Edit:
I wasn’t able to think of a direct approach so I assumed it to be a equilateral Triangle (we get some few inequalities by this method) but that didn’t work ...

Let $$f(x) = \frac{\sin x}{x}, \quad x \in \left[0, \frac{\pi}{2}\right].$$
We have
$$f'(x) = \frac{x \cos x - \sin x}{x^2} = \frac{g(x)}{x^2},$$
and
$$g'(x)= - x\sin x < 0.$$
So $g(x) < g(0) = 0,$ deduced $f'(x) < 0,$ or
$$f(x) > f\left(\frac{\pi}{2}\right) = \frac{2}{\pi}.$$
Therefore
$$\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} > \frac{2}{\pi}+\frac{2}{\pi}+\frac{2}{\pi} = \frac{6}{\pi}.$$
