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 ...
On
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 $$\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} > \frac{2}{\pi}+\frac{2}{\pi}+\frac{2}{\pi} = \frac{6}{\pi}.$$
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We can effectively use the graph of this function for this problem.
The graph of $\frac{\sin x}{x}$ for $x$ belongs to $[0,\pi]$
Assume the three points $(A,\frac{\sin A}{A}),(B,\frac{\sin B}{B}),(C,\frac{\sin C}{C})$ on the graph. The centroid of this triangle is:$$\left(\frac{\left(A+B+C\right)}{3},\frac{\left(\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}\right)}{3}\right)$$
We know that $A+B+C=\pi$ so, the Centroid lies on the line $x=\frac{\pi}{3}$
Graphically,
The centroid lies on the blue line. Also, as the graph is concave down, So the triangleformed always remains below the curve. Hence the centroid always remains below the curve. Its maximum y-coordinate will be when the centroid lies exactly on the curve.
For this to happen, all three points must coincide at $\left(\frac{\pi}{3},\frac{\sin\left(\frac{\pi}{3}\right)}{3}\right)$
So your maximum occurs when $A=B=C=\frac{\pi}{3}$. Hence: $$(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} )> \frac{6}{\pi}$$
It's actually quite simple. The sinc function $\frac{\sin x}x$ is monotonically decreasing and positive on $[0,\pi/2]$; going from $1$ at $x=0$ to $\frac2\pi$ at $x=\frac\pi2$. So $\frac{\sin x}x>\frac2\pi$ for each angle of the acute triangle, and the sum must be strictly greater than $\frac6\pi$.